Solve (x-3)(2x-1)=4*(x+1)^2-x+2

$$( x - 3 ) ( 2 x - 1 ) = 4 \cdot ( x + 1 ) ^ { 2 } - x + 2$$

Solve for x

$$x=\frac{\sqrt{43}-7}{2}\approx -0.221280738$$

$$x=\frac{-\sqrt{43}-7}{2}\approx -6.778719262$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-3 by 2x-1 and combine like terms.

$$2x^{2}-7x+3=4\left(x+1\right)^{2}-x+2$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-7x+3=4\left(x^{2}+2x+1\right)-x+2$$

Use the distributive property to multiply 4 by x^{2}+2x+1.

$$2x^{2}-7x+3=4x^{2}+8x+4-x+2$$

Combine 8x and -x to get 7x.

$$2x^{2}-7x+3=4x^{2}+7x+4+2$$

Add 4 and 2 to get 6.

$$2x^{2}-7x+3=4x^{2}+7x+6$$

Subtract 4x^{2} from both sides.

$$2x^{2}-7x+3-4x^{2}=7x+6$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}-7x+3=7x+6$$

Subtract 7x from both sides.

$$-2x^{2}-7x+3-7x=6$$

Combine -7x and -7x to get -14x.

$$-2x^{2}-14x+3=6$$

Subtract 6 from both sides.

$$-2x^{2}-14x+3-6=0$$

Subtract 6 from 3 to get -3.

$$-2x^{2}-14x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -14 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-14\right)±\sqrt{196+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-\left(-14\right)±\sqrt{196-24}}{2\left(-2\right)}$$

Add 196 to -24.

$$x=\frac{-\left(-14\right)±\sqrt{172}}{2\left(-2\right)}$$

Take the square root of 172.

$$x=\frac{-\left(-14\right)±2\sqrt{43}}{2\left(-2\right)}$$

The opposite of -14 is 14.

$$x=\frac{14±2\sqrt{43}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{14±2\sqrt{43}}{-4}$$

Now solve the equation x=\frac{14±2\sqrt{43}}{-4} when ± is plus. Add 14 to 2\sqrt{43}.

$$x=\frac{2\sqrt{43}+14}{-4}$$

Divide 14+2\sqrt{43} by -4.

$$x=\frac{-\sqrt{43}-7}{2}$$

Now solve the equation x=\frac{14±2\sqrt{43}}{-4} when ± is minus. Subtract 2\sqrt{43} from 14.

$$x=\frac{14-2\sqrt{43}}{-4}$$

Divide 14-2\sqrt{43} by -4.

$$x=\frac{\sqrt{43}-7}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{43}-7}{2} x=\frac{\sqrt{43}-7}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x-3 by 2x-1 and combine like terms.

$$2x^{2}-7x+3=4\left(x+1\right)^{2}-x+2$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-7x+3=4\left(x^{2}+2x+1\right)-x+2$$

Use the distributive property to multiply 4 by x^{2}+2x+1.

$$2x^{2}-7x+3=4x^{2}+8x+4-x+2$$

Combine 8x and -x to get 7x.

$$2x^{2}-7x+3=4x^{2}+7x+4+2$$

Add 4 and 2 to get 6.

$$2x^{2}-7x+3=4x^{2}+7x+6$$

Subtract 4x^{2} from both sides.

$$2x^{2}-7x+3-4x^{2}=7x+6$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}-7x+3=7x+6$$

Subtract 7x from both sides.

$$-2x^{2}-7x+3-7x=6$$

Combine -7x and -7x to get -14x.

$$-2x^{2}-14x+3=6$$

Subtract 3 from both sides.

$$-2x^{2}-14x=6-3$$

Subtract 3 from 6 to get 3.

$$-2x^{2}-14x=3$$

Divide both sides by -2.

$$\frac{-2x^{2}-14x}{-2}=\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{14}{-2}\right)x=\frac{3}{-2}$$

Divide -14 by -2.

$$x^{2}+7x=\frac{3}{-2}$$

Divide 3 by -2.

$$x^{2}+7x=-\frac{3}{2}$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-\frac{3}{2}+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=-\frac{3}{2}+\frac{49}{4}$$

Add -\frac{3}{2} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+7x+\frac{49}{4}=\frac{43}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{43}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{43}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{\sqrt{43}}{2} x+\frac{7}{2}=-\frac{\sqrt{43}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=\frac{\sqrt{43}-7}{2} x=\frac{-\sqrt{43}-7}{2}$$