Solve (10x+8)^2=8/15(120x^2-120x+100)

$$( 10 x + 8 ) ^ { 2 } = \frac { 8 } { 15 } ( 120 x ^ { 2 } - 120 x + 100 )$$

Solve for x

$$x=\frac{2\sqrt{190}-28}{9}\approx -0.047989166$$

$$x=\frac{-2\sqrt{190}-28}{9}\approx -6.174233056$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10x+8\right)^{2}.

$$100x^{2}+160x+64=\frac{8}{15}\left(120x^{2}-120x+100\right)$$

Use the distributive property to multiply \frac{8}{15} by 120x^{2}-120x+100.

$$100x^{2}+160x+64=64x^{2}-64x+\frac{160}{3}$$

Subtract 64x^{2} from both sides.

$$100x^{2}+160x+64-64x^{2}=-64x+\frac{160}{3}$$

Combine 100x^{2} and -64x^{2} to get 36x^{2}.

$$36x^{2}+160x+64=-64x+\frac{160}{3}$$

Add 64x to both sides.

$$36x^{2}+160x+64+64x=\frac{160}{3}$$

Combine 160x and 64x to get 224x.

$$36x^{2}+224x+64=\frac{160}{3}$$

Subtract \frac{160}{3} from both sides.

$$36x^{2}+224x+64-\frac{160}{3}=0$$

Subtract \frac{160}{3} from 64 to get \frac{32}{3}.

$$36x^{2}+224x+\frac{32}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 224 for b, and \frac{32}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-224±\sqrt{224^{2}-4\times 36\times \frac{32}{3}}}{2\times 36}$$

Square 224.

$$x=\frac{-224±\sqrt{50176-4\times 36\times \frac{32}{3}}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-224±\sqrt{50176-144\times \frac{32}{3}}}{2\times 36}$$

Multiply -144 times \frac{32}{3}.

$$x=\frac{-224±\sqrt{50176-1536}}{2\times 36}$$

Add 50176 to -1536.

$$x=\frac{-224±\sqrt{48640}}{2\times 36}$$

Take the square root of 48640.

$$x=\frac{-224±16\sqrt{190}}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-224±16\sqrt{190}}{72}$$

Now solve the equation x=\frac{-224±16\sqrt{190}}{72} when ± is plus. Add -224 to 16\sqrt{190}.

$$x=\frac{16\sqrt{190}-224}{72}$$

Divide -224+16\sqrt{190} by 72.

$$x=\frac{2\sqrt{190}-28}{9}$$

Now solve the equation x=\frac{-224±16\sqrt{190}}{72} when ± is minus. Subtract 16\sqrt{190} from -224.

$$x=\frac{-16\sqrt{190}-224}{72}$$

Divide -224-16\sqrt{190} by 72.

$$x=\frac{-2\sqrt{190}-28}{9}$$

The equation is now solved.

$$x=\frac{2\sqrt{190}-28}{9} x=\frac{-2\sqrt{190}-28}{9}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10x+8\right)^{2}.

$$100x^{2}+160x+64=\frac{8}{15}\left(120x^{2}-120x+100\right)$$

Use the distributive property to multiply \frac{8}{15} by 120x^{2}-120x+100.

$$100x^{2}+160x+64=64x^{2}-64x+\frac{160}{3}$$

Subtract 64x^{2} from both sides.

$$100x^{2}+160x+64-64x^{2}=-64x+\frac{160}{3}$$

Combine 100x^{2} and -64x^{2} to get 36x^{2}.

$$36x^{2}+160x+64=-64x+\frac{160}{3}$$

Add 64x to both sides.

$$36x^{2}+160x+64+64x=\frac{160}{3}$$

Combine 160x and 64x to get 224x.

$$36x^{2}+224x+64=\frac{160}{3}$$

Subtract 64 from both sides.

$$36x^{2}+224x=\frac{160}{3}-64$$

Subtract 64 from \frac{160}{3} to get -\frac{32}{3}.

$$36x^{2}+224x=-\frac{32}{3}$$

Divide both sides by 36.

$$\frac{36x^{2}+224x}{36}=-\frac{\frac{32}{3}}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{224}{36}x=-\frac{\frac{32}{3}}{36}$$

Reduce the fraction \frac{224}{36} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{56}{9}x=-\frac{\frac{32}{3}}{36}$$

Divide -\frac{32}{3} by 36.

$$x^{2}+\frac{56}{9}x=-\frac{8}{27}$$

Divide \frac{56}{9}, the coefficient of the x term, by 2 to get \frac{28}{9}. Then add the square of \frac{28}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{56}{9}x+\left(\frac{28}{9}\right)^{2}=-\frac{8}{27}+\left(\frac{28}{9}\right)^{2}$$

Square \frac{28}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{56}{9}x+\frac{784}{81}=-\frac{8}{27}+\frac{784}{81}$$

Add -\frac{8}{27} to \frac{784}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{56}{9}x+\frac{784}{81}=\frac{760}{81}$$

Factor x^{2}+\frac{56}{9}x+\frac{784}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{28}{9}\right)^{2}=\frac{760}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{28}{9}\right)^{2}}=\sqrt{\frac{760}{81}}$$

Simplify.

$$x+\frac{28}{9}=\frac{2\sqrt{190}}{9} x+\frac{28}{9}=-\frac{2\sqrt{190}}{9}$$

Subtract \frac{28}{9} from both sides of the equation.

$$x=\frac{2\sqrt{190}-28}{9} x=\frac{-2\sqrt{190}-28}{9}$$