Solve 8x^2-80x=-192

$$8 x ^ { 2 } - 80 x = - 192$$

Solve for x

$$x=4$$

$$x=6$$

Steps Using the Quadratic Formula

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.

$$8x^{2}-80x=-192$$

Add 192 to both sides of the equation.

$$8x^{2}-80x-\left(-192\right)=-192-\left(-192\right)$$

Subtracting -192 from itself leaves 0.

$$8x^{2}-80x-\left(-192\right)=0$$

Subtract -192 from 0.

$$8x^{2}-80x+192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -80 for b, and 192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 8\times 192}}{2\times 8}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 8\times 192}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-80\right)±\sqrt{6400-32\times 192}}{2\times 8}$$

Multiply -32 times 192.

$$x=\frac{-\left(-80\right)±\sqrt{6400-6144}}{2\times 8}$$

Add 6400 to -6144.

$$x=\frac{-\left(-80\right)±\sqrt{256}}{2\times 8}$$

Take the square root of 256.

$$x=\frac{-\left(-80\right)±16}{2\times 8}$$

The opposite of -80 is 80.

$$x=\frac{80±16}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{80±16}{16}$$

Now solve the equation x=\frac{80±16}{16} when ± is plus. Add 80 to 16.

$$x=\frac{96}{16}$$

Divide 96 by 16.

$$x=6$$

Now solve the equation x=\frac{80±16}{16} when ± is minus. Subtract 16 from 80.

$$x=\frac{64}{16}$$

Divide 64 by 16.

$$x=4$$

The equation is now solved.

$$x=6 x=4$$

Steps for Completing the Square

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.

$$8x^{2}-80x=-192$$

Divide both sides by 8.

$$\frac{8x^{2}-80x}{8}=-\frac{192}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{80}{8}\right)x=-\frac{192}{8}$$

Divide -80 by 8.

$$x^{2}-10x=-\frac{192}{8}$$

Divide -192 by 8.

$$x^{2}-10x=-24$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-24+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-24+25$$

Add -24 to 25.

$$x^{2}-10x+25=1$$

Factor x^{2}-10x+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}.

$$\left(x-5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-5=1 x-5=-1$$

Add 5 to both sides of the equation.

$$x=6 x=4$$