Solve left(2x-3right)left(3x+1right)+xleft(5-2xright)=7x

$$\left( 2x-3 \right) \left( 3x+1 \right) +x \left( 5-2x \right) = 7x$$

Solve for x

$$x = \frac{\sqrt{129} + 9}{8} \approx 2.544727086$$

$$x=\frac{9-\sqrt{129}}{8}\approx -0.294727086$$

Steps Using the Quadratic Formula

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

$$6x^{2}-7x-3+x\left(5-2x\right)=7x$$

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

$$6x^{2}-7x-3+5x-2x^{2}=7x$$

Combine -7x and 5x to get -2x.

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

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

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

Subtract 7x from both sides.

$$4x^{2}-2x-3-7x=0$$

Combine -2x and -7x to get -9x.

$$4x^{2}-9x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -9 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-9\right)±\sqrt{81-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-\left(-9\right)±\sqrt{81+48}}{2\times 4}$$

Add 81 to 48.

$$x=\frac{-\left(-9\right)±\sqrt{129}}{2\times 4}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{129}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{9±\sqrt{129}}{8}$$

Now solve the equation x=\frac{9±\sqrt{129}}{8} when ± is plus. Add 9 to \sqrt{129}.

$$x=\frac{\sqrt{129}+9}{8}$$

Now solve the equation x=\frac{9±\sqrt{129}}{8} when ± is minus. Subtract \sqrt{129} from 9.

$$x=\frac{9-\sqrt{129}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{129}+9}{8} x=\frac{9-\sqrt{129}}{8}$$

Steps for Completing the Square

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

$$6x^{2}-7x-3+x\left(5-2x\right)=7x$$

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

$$6x^{2}-7x-3+5x-2x^{2}=7x$$

Combine -7x and 5x to get -2x.

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

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

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

Subtract 7x from both sides.

$$4x^{2}-2x-3-7x=0$$

Combine -2x and -7x to get -9x.

$$4x^{2}-9x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$4x^{2}-9x=3$$

Divide both sides by 4.

$$\frac{4x^{2}-9x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{9}{4}x=\frac{3}{4}$$

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{4}x+\left(-\frac{9}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=\frac{3}{4}+\frac{81}{64}$$

Add \frac{3}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=\frac{129}{64}$$

Factor x^{2}-\frac{9}{4}x+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{8}\right)^{2}=\frac{129}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{8}\right)^{2}}=\sqrt{\frac{129}{64}}$$

Simplify.

$$x-\frac{9}{8}=\frac{\sqrt{129}}{8} x-\frac{9}{8}=-\frac{\sqrt{129}}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$x=\frac{\sqrt{129}+9}{8} x=\frac{9-\sqrt{129}}{8}$$