To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

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Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

2x2-x-3=0 Two solutions were found : x = -1 x = 3/2 = 1.500 Step by step solution : Step 1 :Equation at the end of step 1 : (2x2 - x) - 3 = 0 Step 2 :Trying to factor by splitting the ...

2x(2)=19x+33 Two solutions were found : x = -3/2 = -1.500 x = 11 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : ...

2x2=-8x+45 Two solutions were found : x =(-8-√424)/4=-2-1/2√ 106 = -7.148 x =(-8+√424)/4=-2+1/2√ 106 = 3.148 Rearrange: Rearrange the equation by subtracting what is to the right of the ...

24x2-x-3=0 Two solutions were found : x = -1/3 = -0.333 x = 3/8 = 0.375 Step by step solution : Step 1 :Equation at the end of step 1 : ((23•3x2) - x) - 3 = 0 Step 2 :Trying to factor ...

2x2-2x-3=0 Two solutions were found : x =(2-√28)/4=(1-√ 7 )/2= -0.823 x =(2+√28)/4=(1+√ 7 )/2= 1.823 Step by step solution : Step 1 :Equation at the end of step 1 : (2x2 - 2x) - 3 = 0 ...

\displaystyle{x}=\frac{{{3}\pm\sqrt{{{15}}}}}{{2}} Explanation: \displaystyle{2}{x}^{{2}}-{3}{x}-{3}={0} divide by\displaystyle{2}to make coefficient\displaystyle{x}^{{2}}={1} ...

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To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.