Solve 2x(5x-3)=2(4x+3)

$$2x(5x-3)=2(4x+3)$$

Solve for x

$$x = \frac{\sqrt{109} + 7}{10} \approx 1.744030651$$

$$x=\frac{7-\sqrt{109}}{10}\approx -0.344030651$$

Steps Using the Quadratic Formula

Cancel out 2 on both sides.

$$x\left(5x-3\right)=4x+3$$

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

$$5x^{2}-3x=4x+3$$

Subtract 4x from both sides.

$$5x^{2}-3x-4x=3$$

Combine -3x and -4x to get -7x.

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

Subtract 3 from both sides.

$$5x^{2}-7x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-7\right)±\sqrt{49-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-\left(-7\right)±\sqrt{49+60}}{2\times 5}$$

Add 49 to 60.

$$x=\frac{-\left(-7\right)±\sqrt{109}}{2\times 5}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{109}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{7±\sqrt{109}}{10}$$

Now solve the equation x=\frac{7±\sqrt{109}}{10} when ± is plus. Add 7 to \sqrt{109}.

$$x=\frac{\sqrt{109}+7}{10}$$

Now solve the equation x=\frac{7±\sqrt{109}}{10} when ± is minus. Subtract \sqrt{109} from 7.

$$x=\frac{7-\sqrt{109}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{109}+7}{10} x=\frac{7-\sqrt{109}}{10}$$

Steps for Completing the Square

Cancel out 2 on both sides.

$$x\left(5x-3\right)=4x+3$$

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

$$5x^{2}-3x=4x+3$$

Subtract 4x from both sides.

$$5x^{2}-3x-4x=3$$

Combine -3x and -4x to get -7x.

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

Divide both sides by 5.

$$\frac{5x^{2}-7x}{5}=\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{7}{5}x=\frac{3}{5}$$

Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{5}x+\left(-\frac{7}{10}\right)^{2}=\frac{3}{5}+\left(-\frac{7}{10}\right)^{2}$$

Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{3}{5}+\frac{49}{100}$$

Add \frac{3}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{109}{100}$$

Factor x^{2}-\frac{7}{5}x+\frac{49}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{10}\right)^{2}=\frac{109}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{10}\right)^{2}}=\sqrt{\frac{109}{100}}$$

Simplify.

$$x-\frac{7}{10}=\frac{\sqrt{109}}{10} x-\frac{7}{10}=-\frac{\sqrt{109}}{10}$$

Add \frac{7}{10} to both sides of the equation.

$$x=\frac{\sqrt{109}+7}{10} x=\frac{7-\sqrt{109}}{10}$$