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

$$4 \times ( x + 3 ) - 2 x ( x - 1 ) = 3 x - 3 ( 2 x - 1 )$$

Solve for x

$$x = \frac{3 \sqrt{17} + 9}{4} \approx 5.342329219$$

$$x=\frac{9-3\sqrt{17}}{4}\approx -0.842329219$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x+3.

$$4x+12-2x\left(x-1\right)=3x-3\left(2x-1\right)$$

Use the distributive property to multiply -3 by 2x-1.

$$4x+12-2x\left(x-1\right)=3x-6x+3$$

Combine 3x and -6x to get -3x.

$$4x+12-2x\left(x-1\right)=-3x+3$$

Add 3x to both sides.

$$4x+12-2x\left(x-1\right)+3x=3$$

Subtract 3 from both sides.

$$4x+12-2x\left(x-1\right)+3x-3=0$$

Use the distributive property to multiply -2x by x-1.

$$4x+12-2x^{2}+2x+3x-3=0$$

Combine 4x and 2x to get 6x.

$$6x+12-2x^{2}+3x-3=0$$

Combine 6x and 3x to get 9x.

$$9x+12-2x^{2}-3=0$$

Subtract 3 from 12 to get 9.

$$9x+9-2x^{2}=0$$

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.

$$-2x^{2}+9x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 9 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-9±\sqrt{81+8\times 9}}{2\left(-2\right)}$$

Multiply 8 times 9.

$$x=\frac{-9±\sqrt{81+72}}{2\left(-2\right)}$$

Add 81 to 72.

$$x=\frac{-9±\sqrt{153}}{2\left(-2\right)}$$

Take the square root of 153.

$$x=\frac{-9±3\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-9±3\sqrt{17}}{-4}$$

Now solve the equation x=\frac{-9±3\sqrt{17}}{-4} when ± is plus. Add -9 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}-9}{-4}$$

Divide -9+3\sqrt{17} by -4.

$$x=\frac{9-3\sqrt{17}}{4}$$

Now solve the equation x=\frac{-9±3\sqrt{17}}{-4} when ± is minus. Subtract 3\sqrt{17} from -9.

$$x=\frac{-3\sqrt{17}-9}{-4}$$

Divide -9-3\sqrt{17} by -4.

$$x=\frac{3\sqrt{17}+9}{4}$$

The equation is now solved.

$$x=\frac{9-3\sqrt{17}}{4} x=\frac{3\sqrt{17}+9}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 4 by x+3.

$$4x+12-2x\left(x-1\right)=3x-3\left(2x-1\right)$$

Use the distributive property to multiply -3 by 2x-1.

$$4x+12-2x\left(x-1\right)=3x-6x+3$$

Combine 3x and -6x to get -3x.

$$4x+12-2x\left(x-1\right)=-3x+3$$

Add 3x to both sides.

$$4x+12-2x\left(x-1\right)+3x=3$$

Use the distributive property to multiply -2x by x-1.

$$4x+12-2x^{2}+2x+3x=3$$

Combine 4x and 2x to get 6x.

$$6x+12-2x^{2}+3x=3$$

Combine 6x and 3x to get 9x.

$$9x+12-2x^{2}=3$$

Subtract 12 from both sides.

$$9x-2x^{2}=3-12$$

Subtract 12 from 3 to get -9.

$$9x-2x^{2}=-9$$

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.

$$-2x^{2}+9x=-9$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 9 by -2.

$$x^{2}-\frac{9}{2}x=-\frac{9}{-2}$$

Divide -9 by -2.

$$x^{2}-\frac{9}{2}x=\frac{9}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{9}{2}+\frac{81}{16}$$

Add \frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{153}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{153}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{153}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{3\sqrt{17}}{4} x-\frac{9}{4}=-\frac{3\sqrt{17}}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{3\sqrt{17}+9}{4} x=\frac{9-3\sqrt{17}}{4}$$