Solve -5x^2+5=24x

$$- 5 x ^ { 2 } + 5 = 24 x$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=-5$$

Steps Using Factoring By Grouping

Subtract 24x from both sides.

$$-5x^{2}+5-24x=0$$

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

$$-5x^{2}-24x+5=0$$

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

$$a+b=-24 ab=-5\times 5=-25$$

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 -25.

$$1,-25 5,-5$$

Calculate the sum for each pair.

$$1-25=-24 5-5=0$$

The solution is the pair that gives sum -24.

$$a=1 b=-25$$

Rewrite -5x^{2}-24x+5 as \left(-5x^{2}+x\right)+\left(-25x+5\right).

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

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

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

Factor out common term 5x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 24x from both sides.

$$-5x^{2}+5-24x=0$$

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.

$$-5x^{2}-24x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -24 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-24\right)±\sqrt{576+20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$x=\frac{-\left(-24\right)±\sqrt{576+100}}{2\left(-5\right)}$$

Add 576 to 100.

$$x=\frac{-\left(-24\right)±\sqrt{676}}{2\left(-5\right)}$$

Take the square root of 676.

$$x=\frac{-\left(-24\right)±26}{2\left(-5\right)}$$

The opposite of -24 is 24.

$$x=\frac{24±26}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{24±26}{-10}$$

Now solve the equation x=\frac{24±26}{-10} when ± is plus. Add 24 to 26.

$$x=\frac{50}{-10}$$

Divide 50 by -10.

$$x=-5$$

Now solve the equation x=\frac{24±26}{-10} when ± is minus. Subtract 26 from 24.

$$x=-\frac{2}{-10}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 24x from both sides.

$$-5x^{2}+5-24x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}-24x=-5$$

Divide both sides by -5.

$$\frac{-5x^{2}-24x}{-5}=-\frac{5}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{24}{-5}\right)x=-\frac{5}{-5}$$

Divide -24 by -5.

$$x^{2}+\frac{24}{5}x=-\frac{5}{-5}$$

Divide -5 by -5.

$$x^{2}+\frac{24}{5}x=1$$

Divide \frac{24}{5}, the coefficient of the x term, by 2 to get \frac{12}{5}. Then add the square of \frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{24}{5}x+\left(\frac{12}{5}\right)^{2}=1+\left(\frac{12}{5}\right)^{2}$$

Square \frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{24}{5}x+\frac{144}{25}=1+\frac{144}{25}$$

Add 1 to \frac{144}{25}.

$$x^{2}+\frac{24}{5}x+\frac{144}{25}=\frac{169}{25}$$

Factor x^{2}+\frac{24}{5}x+\frac{144}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{12}{5}\right)^{2}=\frac{169}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{12}{5}\right)^{2}}=\sqrt{\frac{169}{25}}$$

Simplify.

$$x+\frac{12}{5}=\frac{13}{5} x+\frac{12}{5}=-\frac{13}{5}$$

Subtract \frac{12}{5} from both sides of the equation.

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