Solve 4-10(x+6)=5(x-1)32x

$$4-10(x+6)=5(x-1)32x$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3335}i}{160}+\frac{15}{32}\approx 0.46875+0.360934118i$$

$$x=-\frac{\sqrt{3335}i}{160}+\frac{15}{32}\approx 0.46875-0.360934118i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -10 by x+6.

$$4-10x-60=5\left(x-1\right)\times 32x$$

Subtract 60 from 4 to get -56.

$$-56-10x=5\left(x-1\right)\times 32x$$

Multiply 5 and 32 to get 160.

$$-56-10x=160\left(x-1\right)x$$

Use the distributive property to multiply 160 by x-1.

$$-56-10x=\left(160x-160\right)x$$

Use the distributive property to multiply 160x-160 by x.

$$-56-10x=160x^{2}-160x$$

Subtract 160x^{2} from both sides.

$$-56-10x-160x^{2}=-160x$$

Add 160x to both sides.

$$-56-10x-160x^{2}+160x=0$$

Combine -10x and 160x to get 150x.

$$-56+150x-160x^{2}=0$$

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.

$$-160x^{2}+150x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -160 for a, 150 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-150±\sqrt{150^{2}-4\left(-160\right)\left(-56\right)}}{2\left(-160\right)}$$

Square 150.

$$x=\frac{-150±\sqrt{22500-4\left(-160\right)\left(-56\right)}}{2\left(-160\right)}$$

Multiply -4 times -160.

$$x=\frac{-150±\sqrt{22500+640\left(-56\right)}}{2\left(-160\right)}$$

Multiply 640 times -56.

$$x=\frac{-150±\sqrt{22500-35840}}{2\left(-160\right)}$$

Add 22500 to -35840.

$$x=\frac{-150±\sqrt{-13340}}{2\left(-160\right)}$$

Take the square root of -13340.

$$x=\frac{-150±2\sqrt{3335}i}{2\left(-160\right)}$$

Multiply 2 times -160.

$$x=\frac{-150±2\sqrt{3335}i}{-320}$$

Now solve the equation x=\frac{-150±2\sqrt{3335}i}{-320} when ± is plus. Add -150 to 2i\sqrt{3335}.

$$x=\frac{-150+2\sqrt{3335}i}{-320}$$

Divide -150+2i\sqrt{3335} by -320.

$$x=-\frac{\sqrt{3335}i}{160}+\frac{15}{32}$$

Now solve the equation x=\frac{-150±2\sqrt{3335}i}{-320} when ± is minus. Subtract 2i\sqrt{3335} from -150.

$$x=\frac{-2\sqrt{3335}i-150}{-320}$$

Divide -150-2i\sqrt{3335} by -320.

$$x=\frac{\sqrt{3335}i}{160}+\frac{15}{32}$$

The equation is now solved.

$$x=-\frac{\sqrt{3335}i}{160}+\frac{15}{32} x=\frac{\sqrt{3335}i}{160}+\frac{15}{32}$$

Steps for Completing the Square

Use the distributive property to multiply -10 by x+6.

$$4-10x-60=5\left(x-1\right)\times 32x$$

Subtract 60 from 4 to get -56.

$$-56-10x=5\left(x-1\right)\times 32x$$

Multiply 5 and 32 to get 160.

$$-56-10x=160\left(x-1\right)x$$

Use the distributive property to multiply 160 by x-1.

$$-56-10x=\left(160x-160\right)x$$

Use the distributive property to multiply 160x-160 by x.

$$-56-10x=160x^{2}-160x$$

Subtract 160x^{2} from both sides.

$$-56-10x-160x^{2}=-160x$$

Add 160x to both sides.

$$-56-10x-160x^{2}+160x=0$$

Combine -10x and 160x to get 150x.

$$-56+150x-160x^{2}=0$$

Add 56 to both sides. Anything plus zero gives itself.

$$150x-160x^{2}=56$$

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.

$$-160x^{2}+150x=56$$

Divide both sides by -160.

$$\frac{-160x^{2}+150x}{-160}=\frac{56}{-160}$$

Dividing by -160 undoes the multiplication by -160.

$$x^{2}+\frac{150}{-160}x=\frac{56}{-160}$$

Reduce the fraction \frac{150}{-160} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{15}{16}x=\frac{56}{-160}$$

Reduce the fraction \frac{56}{-160} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{15}{16}x=-\frac{7}{20}$$

Divide -\frac{15}{16}, the coefficient of the x term, by 2 to get -\frac{15}{32}. Then add the square of -\frac{15}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{16}x+\left(-\frac{15}{32}\right)^{2}=-\frac{7}{20}+\left(-\frac{15}{32}\right)^{2}$$

Square -\frac{15}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{16}x+\frac{225}{1024}=-\frac{7}{20}+\frac{225}{1024}$$

Add -\frac{7}{20} to \frac{225}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{16}x+\frac{225}{1024}=-\frac{667}{5120}$$

Factor x^{2}-\frac{15}{16}x+\frac{225}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{32}\right)^{2}=-\frac{667}{5120}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{32}\right)^{2}}=\sqrt{-\frac{667}{5120}}$$

Simplify.

$$x-\frac{15}{32}=\frac{\sqrt{3335}i}{160} x-\frac{15}{32}=-\frac{\sqrt{3335}i}{160}$$

Add \frac{15}{32} to both sides of the equation.

$$x=\frac{\sqrt{3335}i}{160}+\frac{15}{32} x=-\frac{\sqrt{3335}i}{160}+\frac{15}{32}$$