Solve -20x-25=4x^2

$$- 20 x - 25 = 4 x ^ { 2 }$$

Solve for x

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

Steps Using Factoring By Grouping

Subtract 4x^{2} from both sides.

$$-20x-25-4x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}-20x-25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=-20 ab=-4\left(-25\right)=100$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$a=-10 b=-10$$

Rewrite -4x^{2}-20x-25 as \left(-4x^{2}-10x\right)+\left(-10x-25\right).

$$\left(-4x^{2}-10x\right)+\left(-10x-25\right)$$

Factor out 2x in the first and 5 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$-20x-25-4x^{2}=0$$

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.

$$-4x^{2}-20x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -20 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-4\right)\left(-25\right)}}{2\left(-4\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-4\right)\left(-25\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-20\right)±\sqrt{400+16\left(-25\right)}}{2\left(-4\right)}$$

Multiply 16 times -25.

$$x=\frac{-\left(-20\right)±\sqrt{400-400}}{2\left(-4\right)}$$

Add 400 to -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\left(-4\right)}$$

Take the square root of 0.

$$x=-\frac{-20}{2\left(-4\right)}$$

The opposite of -20 is 20.

$$x=\frac{20}{2\left(-4\right)}$$

Multiply 2 times -4.

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

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

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

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$-20x-25-4x^{2}=0$$

Add 25 to both sides. Anything plus zero gives itself.

$$-20x-4x^{2}=25$$

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.

$$-4x^{2}-20x=25$$

Divide both sides by -4.

$$\frac{-4x^{2}-20x}{-4}=\frac{25}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{20}{-4}\right)x=\frac{25}{-4}$$

Divide -20 by -4.

$$x^{2}+5x=\frac{25}{-4}$$

Divide 25 by -4.

$$x^{2}+5x=-\frac{25}{4}$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=\frac{-25+25}{4}$$

Add -\frac{25}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=0$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{5}{2}=0 x+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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