Solve 4(5)(x+5)+16(5)(x-5)=4(x^2-25)

$$4 ( 5 ) ( x + 5 ) + 16 ( 5 ) ( x - 5 ) = 4 ( x ^ { 2 } - 25 )$$

Solve for x

$$x = \frac{5 \sqrt{17} + 25}{2} \approx 22.807764064$$

$$x = \frac{25 - 5 \sqrt{17}}{2} \approx 2.192235936$$

Steps Using the Quadratic Formula

Do the multiplications.

$$20\left(x+5\right)+80\left(x-5\right)=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 20 by x+5.

$$20x+100+80\left(x-5\right)=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 80 by x-5.

$$20x+100+80x-400=4\left(x^{2}-25\right)$$

Combine 20x and 80x to get 100x.

$$100x+100-400=4\left(x^{2}-25\right)$$

Subtract 400 from 100 to get -300.

$$100x-300=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 4 by x^{2}-25.

$$100x-300=4x^{2}-100$$

Subtract 4x^{2} from both sides.

$$100x-300-4x^{2}=-100$$

Add 100 to both sides.

$$100x-300-4x^{2}+100=0$$

Add -300 and 100 to get -200.

$$100x-200-4x^{2}=0$$

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.

$$-4x^{2}+100x-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 100 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-4\right)\left(-200\right)}}{2\left(-4\right)}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-4\right)\left(-200\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-100±\sqrt{10000+16\left(-200\right)}}{2\left(-4\right)}$$

Multiply 16 times -200.

$$x=\frac{-100±\sqrt{10000-3200}}{2\left(-4\right)}$$

Add 10000 to -3200.

$$x=\frac{-100±\sqrt{6800}}{2\left(-4\right)}$$

Take the square root of 6800.

$$x=\frac{-100±20\sqrt{17}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-100±20\sqrt{17}}{-8}$$

Now solve the equation x=\frac{-100±20\sqrt{17}}{-8} when ± is plus. Add -100 to 20\sqrt{17}.

$$x=\frac{20\sqrt{17}-100}{-8}$$

Divide -100+20\sqrt{17} by -8.

$$x=\frac{25-5\sqrt{17}}{2}$$

Now solve the equation x=\frac{-100±20\sqrt{17}}{-8} when ± is minus. Subtract 20\sqrt{17} from -100.

$$x=\frac{-20\sqrt{17}-100}{-8}$$

Divide -100-20\sqrt{17} by -8.

$$x=\frac{5\sqrt{17}+25}{2}$$

The equation is now solved.

$$x=\frac{25-5\sqrt{17}}{2} x=\frac{5\sqrt{17}+25}{2}$$

Steps for Completing the Square

Do the multiplications.

$$20\left(x+5\right)+80\left(x-5\right)=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 20 by x+5.

$$20x+100+80\left(x-5\right)=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 80 by x-5.

$$20x+100+80x-400=4\left(x^{2}-25\right)$$

Combine 20x and 80x to get 100x.

$$100x+100-400=4\left(x^{2}-25\right)$$

Subtract 400 from 100 to get -300.

$$100x-300=4\left(x^{2}-25\right)$$

Use the distributive property to multiply 4 by x^{2}-25.

$$100x-300=4x^{2}-100$$

Subtract 4x^{2} from both sides.

$$100x-300-4x^{2}=-100$$

Add 300 to both sides.

$$100x-4x^{2}=-100+300$$

Add -100 and 300 to get 200.

$$100x-4x^{2}=200$$

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.

$$-4x^{2}+100x=200$$

Divide both sides by -4.

$$\frac{-4x^{2}+100x}{-4}=\frac{200}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{100}{-4}x=\frac{200}{-4}$$

Divide 100 by -4.

$$x^{2}-25x=\frac{200}{-4}$$

Divide 200 by -4.

$$x^{2}-25x=-50$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-50+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-50+\frac{625}{4}$$

Add -50 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=\frac{425}{4}$$

Factor x^{2}-25x+\frac{625}{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}.

$$\left(x-\frac{25}{2}\right)^{2}=\frac{425}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{\frac{425}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{5\sqrt{17}}{2} x-\frac{25}{2}=-\frac{5\sqrt{17}}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{5\sqrt{17}+25}{2} x=\frac{25-5\sqrt{17}}{2}$$