Solve 2left(x-3xleft(5-2xright)right)=10left(-x+9right)

$$2 \left( x-3x \left( 5-2x \right) \right) = 10 \left( -x+9 \right)$$

Solve for x

$$x = \frac{\sqrt{129} + 3}{4} \approx 3.589454173$$

$$x=\frac{3-\sqrt{129}}{4}\approx -2.089454173$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 10 by -x+9.

$$2\left(x-3x\left(5-2x\right)\right)=10\left(-x\right)+90$$

Subtract 10\left(-x\right) from both sides.

$$2\left(x-3x\left(5-2x\right)\right)-10\left(-x\right)=90$$

Subtract 90 from both sides.

$$2\left(x-3x\left(5-2x\right)\right)-10\left(-x\right)-90=0$$

Multiply -1 and 3 to get -3.

$$2\left(x-3x\left(5-2x\right)\right)-10\left(-1\right)x-90=0$$

Use the distributive property to multiply -3x by 5-2x.

$$2\left(x-15x+6x^{2}\right)-10\left(-1\right)x-90=0$$

Combine x and -15x to get -14x.

$$2\left(-14x+6x^{2}\right)-10\left(-1\right)x-90=0$$

Use the distributive property to multiply 2 by -14x+6x^{2}.

$$-28x+12x^{2}-10\left(-1\right)x-90=0$$

Multiply -10 and -1 to get 10.

$$-28x+12x^{2}+10x-90=0$$

Combine -28x and 10x to get -18x.

$$-18x+12x^{2}-90=0$$

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.

$$12x^{2}-18x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -18 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 12\left(-90\right)}}{2\times 12}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 12\left(-90\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-18\right)±\sqrt{324-48\left(-90\right)}}{2\times 12}$$

Multiply -48 times -90.

$$x=\frac{-\left(-18\right)±\sqrt{324+4320}}{2\times 12}$$

Add 324 to 4320.

$$x=\frac{-\left(-18\right)±\sqrt{4644}}{2\times 12}$$

Take the square root of 4644.

$$x=\frac{-\left(-18\right)±6\sqrt{129}}{2\times 12}$$

The opposite of -18 is 18.

$$x=\frac{18±6\sqrt{129}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{18±6\sqrt{129}}{24}$$

Now solve the equation x=\frac{18±6\sqrt{129}}{24} when ± is plus. Add 18 to 6\sqrt{129}.

$$x=\frac{6\sqrt{129}+18}{24}$$

Divide 18+6\sqrt{129} by 24.

$$x=\frac{\sqrt{129}+3}{4}$$

Now solve the equation x=\frac{18±6\sqrt{129}}{24} when ± is minus. Subtract 6\sqrt{129} from 18.

$$x=\frac{18-6\sqrt{129}}{24}$$

Divide 18-6\sqrt{129} by 24.

$$x=\frac{3-\sqrt{129}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{129}+3}{4} x=\frac{3-\sqrt{129}}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 10 by -x+9.

$$2\left(x-3x\left(5-2x\right)\right)=10\left(-x\right)+90$$

Subtract 10\left(-x\right) from both sides.

$$2\left(x-3x\left(5-2x\right)\right)-10\left(-x\right)=90$$

Multiply -1 and 3 to get -3.

$$2\left(x-3x\left(5-2x\right)\right)-10\left(-1\right)x=90$$

Use the distributive property to multiply -3x by 5-2x.

$$2\left(x-15x+6x^{2}\right)-10\left(-1\right)x=90$$

Combine x and -15x to get -14x.

$$2\left(-14x+6x^{2}\right)-10\left(-1\right)x=90$$

Use the distributive property to multiply 2 by -14x+6x^{2}.

$$-28x+12x^{2}-10\left(-1\right)x=90$$

Multiply -10 and -1 to get 10.

$$-28x+12x^{2}+10x=90$$

Combine -28x and 10x to get -18x.

$$-18x+12x^{2}=90$$

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.

$$12x^{2}-18x=90$$

Divide both sides by 12.

$$\frac{12x^{2}-18x}{12}=\frac{90}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{18}{12}\right)x=\frac{90}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{3}{2}x=\frac{90}{12}$$

Reduce the fraction \frac{90}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{3}{2}x=\frac{15}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{15}{2}+\frac{9}{16}$$

Add \frac{15}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{129}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{129}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{129}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{129}}{4} x-\frac{3}{4}=-\frac{\sqrt{129}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{129}+3}{4} x=\frac{3-\sqrt{129}}{4}$$