Solve 5x^2-125=(4x+3)(2x-10)

$$5 x ^ { 2 } - 125 = ( 4 x + 3 ) ( 2 x - 10 )$$

Solve for x

$$x=5$$

$$x = \frac{19}{3} = 6\frac{1}{3} \approx 6.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 4x+3 by 2x-10 and combine like terms.

$$5x^{2}-125=8x^{2}-34x-30$$

Subtract 8x^{2} from both sides.

$$5x^{2}-125-8x^{2}=-34x-30$$

Combine 5x^{2} and -8x^{2} to get -3x^{2}.

$$-3x^{2}-125=-34x-30$$

Add 34x to both sides.

$$-3x^{2}-125+34x=-30$$

Add 30 to both sides.

$$-3x^{2}-125+34x+30=0$$

Add -125 and 30 to get -95.

$$-3x^{2}-95+34x=0$$

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

$$-3x^{2}+34x-95=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-95. To find a and b, set up a system to be solved.

$$a+b=34 ab=-3\left(-95\right)=285$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 285.

$$1,285 3,95 5,57 15,19$$

Calculate the sum for each pair.

$$1+285=286 3+95=98 5+57=62 15+19=34$$

The solution is the pair that gives sum 34.

$$a=19 b=15$$

Rewrite -3x^{2}+34x-95 as \left(-3x^{2}+19x\right)+\left(15x-95\right).

$$\left(-3x^{2}+19x\right)+\left(15x-95\right)$$

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

$$-x\left(3x-19\right)+5\left(3x-19\right)$$

Factor out common term 3x-19 by using distributive property.

$$\left(3x-19\right)\left(-x+5\right)$$

To find equation solutions, solve 3x-19=0 and -x+5=0.

$$x=\frac{19}{3} x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x+3 by 2x-10 and combine like terms.

$$5x^{2}-125=8x^{2}-34x-30$$

Subtract 8x^{2} from both sides.

$$5x^{2}-125-8x^{2}=-34x-30$$

Combine 5x^{2} and -8x^{2} to get -3x^{2}.

$$-3x^{2}-125=-34x-30$$

Add 34x to both sides.

$$-3x^{2}-125+34x=-30$$

Add 30 to both sides.

$$-3x^{2}-125+34x+30=0$$

Add -125 and 30 to get -95.

$$-3x^{2}-95+34x=0$$

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.

$$-3x^{2}+34x-95=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 34 for b, and -95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-34±\sqrt{34^{2}-4\left(-3\right)\left(-95\right)}}{2\left(-3\right)}$$

Square 34.

$$x=\frac{-34±\sqrt{1156-4\left(-3\right)\left(-95\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-34±\sqrt{1156+12\left(-95\right)}}{2\left(-3\right)}$$

Multiply 12 times -95.

$$x=\frac{-34±\sqrt{1156-1140}}{2\left(-3\right)}$$

Add 1156 to -1140.

$$x=\frac{-34±\sqrt{16}}{2\left(-3\right)}$$

Take the square root of 16.

$$x=\frac{-34±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-34±4}{-6}$$

Now solve the equation x=\frac{-34±4}{-6} when ± is plus. Add -34 to 4.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

Now solve the equation x=\frac{-34±4}{-6} when ± is minus. Subtract 4 from -34.

$$x=-\frac{38}{-6}$$

Reduce the fraction \frac{-38}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{19}{3}$$

The equation is now solved.

$$x=5 x=\frac{19}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 4x+3 by 2x-10 and combine like terms.

$$5x^{2}-125=8x^{2}-34x-30$$

Subtract 8x^{2} from both sides.

$$5x^{2}-125-8x^{2}=-34x-30$$

Combine 5x^{2} and -8x^{2} to get -3x^{2}.

$$-3x^{2}-125=-34x-30$$

Add 34x to both sides.

$$-3x^{2}-125+34x=-30$$

Add 125 to both sides.

$$-3x^{2}+34x=-30+125$$

Add -30 and 125 to get 95.

$$-3x^{2}+34x=95$$

Divide both sides by -3.

$$\frac{-3x^{2}+34x}{-3}=\frac{95}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{34}{-3}x=\frac{95}{-3}$$

Divide 34 by -3.

$$x^{2}-\frac{34}{3}x=\frac{95}{-3}$$

Divide 95 by -3.

$$x^{2}-\frac{34}{3}x=-\frac{95}{3}$$

Divide -\frac{34}{3}, the coefficient of the x term, by 2 to get -\frac{17}{3}. Then add the square of -\frac{17}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{34}{3}x+\left(-\frac{17}{3}\right)^{2}=-\frac{95}{3}+\left(-\frac{17}{3}\right)^{2}$$

Square -\frac{17}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{34}{3}x+\frac{289}{9}=-\frac{95}{3}+\frac{289}{9}$$

Add -\frac{95}{3} to \frac{289}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{34}{3}x+\frac{289}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{34}{3}x+\frac{289}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x-\frac{17}{3}=\frac{2}{3} x-\frac{17}{3}=-\frac{2}{3}$$

Add \frac{17}{3} to both sides of the equation.

$$x=\frac{19}{3} x=5$$