Solve -[2(2x^2-5x+2x-5)]=0

$$- [ 2 ( 2 x ^ { 2 } - 5 x + 2 x - 5 ) ] = 0$$

Solve for x

$$x=-1$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Combine -5x and 2x to get -3x.

$$-2\left(2x^{2}-3x-5\right)=0$$

Use the distributive property to multiply 2 by 2x^{2}-3x-5.

$$-\left(4x^{2}-6x-10\right)=0$$

To find the opposite of 4x^{2}-6x-10, find the opposite of each term.

$$-4x^{2}+6x+10=0$$

Divide both sides by 2.

$$-2x^{2}+3x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=3 ab=-2\times 5=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=5 b=-2$$

Rewrite -2x^{2}+3x+5 as \left(-2x^{2}+5x\right)+\left(-2x+5\right).

$$\left(-2x^{2}+5x\right)+\left(-2x+5\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(2x-5\right)-\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(-x-1\right)$$

To find equation solutions, solve 2x-5=0 and -x-1=0.

$$x=\frac{5}{2} x=-1$$

Steps Using the Quadratic Formula

Combine -5x and 2x to get -3x.

$$-2\left(2x^{2}-3x-5\right)=0$$

Use the distributive property to multiply 2 by 2x^{2}-3x-5.

$$-\left(4x^{2}-6x-10\right)=0$$

To find the opposite of 4x^{2}-6x-10, find the opposite of each term.

$$-4x^{2}+6x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 6 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)\times 10}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-6±\sqrt{36+16\times 10}}{2\left(-4\right)}$$

Multiply 16 times 10.

$$x=\frac{-6±\sqrt{36+160}}{2\left(-4\right)}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2\left(-4\right)}$$

Take the square root of 196.

$$x=\frac{-6±14}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±14}{-8}$$

Now solve the equation x=\frac{-6±14}{-8} when ± is plus. Add -6 to 14.

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

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-6±14}{-8} when ± is minus. Subtract 14 from -6.

$$x=-\frac{20}{-8}$$

Reduce the fraction \frac{-20}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=-1 x=\frac{5}{2}$$

Steps for Completing the Square

Combine -5x and 2x to get -3x.

$$-2\left(2x^{2}-3x-5\right)=0$$

Use the distributive property to multiply 2 by 2x^{2}-3x-5.

$$-\left(4x^{2}-6x-10\right)=0$$

To find the opposite of 4x^{2}-6x-10, find the opposite of each term.

$$-4x^{2}+6x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}+6x=-10$$

Divide both sides by -4.

$$\frac{-4x^{2}+6x}{-4}=-\frac{10}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{6}{-4}x=-\frac{10}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{2}x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{2}x=\frac{5}{2}$$

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.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{3}{4}\right)^{2}$$

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{7}{4} x-\frac{3}{4}=-\frac{7}{4}$$

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

$$x=\frac{5}{2} x=-1$$