Solve 3x(x-1)+4x=3/4(x+1)-6x

$$3 x ( x - 1 ) + 4 x = \frac { 3 } { 4 } ( x + 1 ) - 6 x$$

Solve for x

$$x=\frac{\sqrt{769}-25}{24}\approx 0.113785385$$

$$x=\frac{-\sqrt{769}-25}{24}\approx -2.197118719$$

Steps Using the Quadratic Formula

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

$$3x^{2}-3x+4x=\frac{3}{4}\left(x+1\right)-6x$$

Combine -3x and 4x to get x.

$$3x^{2}+x=\frac{3}{4}\left(x+1\right)-6x$$

Use the distributive property to multiply \frac{3}{4} by x+1.

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

Combine \frac{3}{4}x and -6x to get -\frac{21}{4}x.

$$3x^{2}+x=-\frac{21}{4}x+\frac{3}{4}$$

Add \frac{21}{4}x to both sides.

$$3x^{2}+x+\frac{21}{4}x=\frac{3}{4}$$

Combine x and \frac{21}{4}x to get \frac{25}{4}x.

$$3x^{2}+\frac{25}{4}x=\frac{3}{4}$$

Subtract \frac{3}{4} from both sides.

$$3x^{2}+\frac{25}{4}x-\frac{3}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, \frac{25}{4} for b, and -\frac{3}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{25}{4}±\sqrt{\left(\frac{25}{4}\right)^{2}-4\times 3\left(-\frac{3}{4}\right)}}{2\times 3}$$

Square \frac{25}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}-4\times 3\left(-\frac{3}{4}\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}-12\left(-\frac{3}{4}\right)}}{2\times 3}$$

Multiply -12 times -\frac{3}{4}.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}+9}}{2\times 3}$$

Add \frac{625}{16} to 9.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{769}{16}}}{2\times 3}$$

Take the square root of \frac{769}{16}.

$$x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6}$$

Now solve the equation x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6} when ± is plus. Add -\frac{25}{4} to \frac{\sqrt{769}}{4}.

$$x=\frac{\sqrt{769}-25}{4\times 6}$$

Divide \frac{-25+\sqrt{769}}{4} by 6.

$$x=\frac{\sqrt{769}-25}{24}$$

Now solve the equation x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6} when ± is minus. Subtract \frac{\sqrt{769}}{4} from -\frac{25}{4}.

$$x=\frac{-\sqrt{769}-25}{4\times 6}$$

Divide \frac{-25-\sqrt{769}}{4} by 6.

$$x=\frac{-\sqrt{769}-25}{24}$$

The equation is now solved.

$$x=\frac{\sqrt{769}-25}{24} x=\frac{-\sqrt{769}-25}{24}$$

Steps for Completing the Square

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

$$3x^{2}-3x+4x=\frac{3}{4}\left(x+1\right)-6x$$

Combine -3x and 4x to get x.

$$3x^{2}+x=\frac{3}{4}\left(x+1\right)-6x$$

Use the distributive property to multiply \frac{3}{4} by x+1.

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

Combine \frac{3}{4}x and -6x to get -\frac{21}{4}x.

$$3x^{2}+x=-\frac{21}{4}x+\frac{3}{4}$$

Add \frac{21}{4}x to both sides.

$$3x^{2}+x+\frac{21}{4}x=\frac{3}{4}$$

Combine x and \frac{21}{4}x to get \frac{25}{4}x.

$$3x^{2}+\frac{25}{4}x=\frac{3}{4}$$

Divide both sides by 3.

$$\frac{3x^{2}+\frac{25}{4}x}{3}=\frac{\frac{3}{4}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{\frac{25}{4}}{3}x=\frac{\frac{3}{4}}{3}$$

Divide \frac{25}{4} by 3.

$$x^{2}+\frac{25}{12}x=\frac{\frac{3}{4}}{3}$$

Divide \frac{3}{4} by 3.

$$x^{2}+\frac{25}{12}x=\frac{1}{4}$$

Divide \frac{25}{12}, the coefficient of the x term, by 2 to get \frac{25}{24}. Then add the square of \frac{25}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{25}{12}x+\left(\frac{25}{24}\right)^{2}=\frac{1}{4}+\left(\frac{25}{24}\right)^{2}$$

Square \frac{25}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{1}{4}+\frac{625}{576}$$

Add \frac{1}{4} to \frac{625}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{769}{576}$$

Factor x^{2}+\frac{25}{12}x+\frac{625}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{25}{24}\right)^{2}=\frac{769}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{24}\right)^{2}}=\sqrt{\frac{769}{576}}$$

Simplify.

$$x+\frac{25}{24}=\frac{\sqrt{769}}{24} x+\frac{25}{24}=-\frac{\sqrt{769}}{24}$$

Subtract \frac{25}{24} from both sides of the equation.

$$x=\frac{\sqrt{769}-25}{24} x=\frac{-\sqrt{769}-25}{24}$$