Solve 2x^2-3x(2x-1)/2=-1

$$2 x ^ { 2 } - \frac { 3 x ( 2 x - 1 ) } { 2 } = - 1$$

Solve for x

$$x=2$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$4x^{2}-3x\left(2x-1\right)=-2$$

Use the distributive property to multiply -3x by 2x-1.

$$4x^{2}-6x^{2}+3x=-2$$

Combine 4x^{2} and -6x^{2} to get -2x^{2}.

$$-2x^{2}+3x=-2$$

Add 2 to both sides.

$$-2x^{2}+3x+2=0$$

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

$$a+b=3 ab=-2\times 2=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -2x^{2}+3x+2 as \left(-2x^{2}+4x\right)+\left(-x+2\right).

$$\left(-2x^{2}+4x\right)+\left(-x+2\right)$$

Factor out 2x in -2x^{2}+4x.

$$2x\left(-x+2\right)-x+2$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(2x+1\right)$$

To find equation solutions, solve -x+2=0 and 2x+1=0.

$$x=2 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$4x^{2}-3x\left(2x-1\right)=-2$$

Use the distributive property to multiply -3x by 2x-1.

$$4x^{2}-6x^{2}+3x=-2$$

Combine 4x^{2} and -6x^{2} to get -2x^{2}.

$$-2x^{2}+3x=-2$$

Add 2 to both sides.

$$-2x^{2}+3x+2=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{-3±\sqrt{9+16}}{2\left(-2\right)}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±5}{-4}$$

Now solve the equation x=\frac{-3±5}{-4} when ± is plus. Add -3 to 5.

$$x=\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-3±5}{-4} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-\frac{1}{2} x=2$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$4x^{2}-3x\left(2x-1\right)=-2$$

Use the distributive property to multiply -3x by 2x-1.

$$4x^{2}-6x^{2}+3x=-2$$

Combine 4x^{2} and -6x^{2} to get -2x^{2}.

$$-2x^{2}+3x=-2$$

Divide both sides by -2.

$$\frac{-2x^{2}+3x}{-2}=-\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{3}{-2}x=-\frac{2}{-2}$$

Divide 3 by -2.

$$x^{2}-\frac{3}{2}x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x^{2}-\frac{3}{2}x=1$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=1+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{5}{4} x-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=2 x=-\frac{1}{2}$$