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

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

Solve for x

$$x=\frac{\sqrt{561}-9}{20}\approx 0.734271928$$

$$x=\frac{-\sqrt{561}-9}{20}\approx -1.634271928$$

Steps Using the Quadratic Formula

Add 2 and 1 to get 3.

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

Use the distributive property to multiply 2x+3 by 5x-3 and combine like terms.

$$3=10x^{2}+9x-9$$

Swap sides so that all variable terms are on the left hand side.

$$10x^{2}+9x-9=3$$

Subtract 3 from both sides.

$$10x^{2}+9x-9-3=0$$

Subtract 3 from -9 to get -12.

$$10x^{2}+9x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 9 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 10\left(-12\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-9±\sqrt{81-40\left(-12\right)}}{2\times 10}$$

Multiply -40 times -12.

$$x=\frac{-9±\sqrt{81+480}}{2\times 10}$$

Add 81 to 480.

$$x=\frac{-9±\sqrt{561}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-9±\sqrt{561}}{20}$$

Now solve the equation x=\frac{-9±\sqrt{561}}{20} when ± is plus. Add -9 to \sqrt{561}.

$$x=\frac{\sqrt{561}-9}{20}$$

Now solve the equation x=\frac{-9±\sqrt{561}}{20} when ± is minus. Subtract \sqrt{561} from -9.

$$x=\frac{-\sqrt{561}-9}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{561}-9}{20} x=\frac{-\sqrt{561}-9}{20}$$

Steps for Completing the Square

Add 2 and 1 to get 3.

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

Use the distributive property to multiply 2x+3 by 5x-3 and combine like terms.

$$3=10x^{2}+9x-9$$

Swap sides so that all variable terms are on the left hand side.

$$10x^{2}+9x-9=3$$

Add 9 to both sides.

$$10x^{2}+9x=3+9$$

Add 3 and 9 to get 12.

$$10x^{2}+9x=12$$

Divide both sides by 10.

$$\frac{10x^{2}+9x}{10}=\frac{12}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{9}{10}x=\frac{12}{10}$$

Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{9}{10}x=\frac{6}{5}$$

Divide \frac{9}{10}, the coefficient of the x term, by 2 to get \frac{9}{20}. Then add the square of \frac{9}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{10}x+\left(\frac{9}{20}\right)^{2}=\frac{6}{5}+\left(\frac{9}{20}\right)^{2}$$

Square \frac{9}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{10}x+\frac{81}{400}=\frac{6}{5}+\frac{81}{400}$$

Add \frac{6}{5} to \frac{81}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{10}x+\frac{81}{400}=\frac{561}{400}$$

Factor x^{2}+\frac{9}{10}x+\frac{81}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{20}\right)^{2}=\frac{561}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{20}\right)^{2}}=\sqrt{\frac{561}{400}}$$

Simplify.

$$x+\frac{9}{20}=\frac{\sqrt{561}}{20} x+\frac{9}{20}=-\frac{\sqrt{561}}{20}$$

Subtract \frac{9}{20} from both sides of the equation.

$$x=\frac{\sqrt{561}-9}{20} x=\frac{-\sqrt{561}-9}{20}$$