Solve left(2x-5right)left(x+3right)=15-6x

$$\left( 2x-5 \right) \left( x+3 \right) = 15-6x$$

Solve for x

$$x=-6$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using the Quadratic Formula

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

$$2x^{2}+x-15=15-6x$$

Subtract 15 from both sides.

$$2x^{2}+x-15-15=-6x$$

Subtract 15 from -15 to get -30.

$$2x^{2}+x-30=-6x$$

Add 6x to both sides.

$$2x^{2}+x-30+6x=0$$

Combine x and 6x to get 7x.

$$2x^{2}+7x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-30\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$x=\frac{-7±\sqrt{49+240}}{2\times 2}$$

Add 49 to 240.

$$x=\frac{-7±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$x=\frac{-7±17}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±17}{4}$$

Now solve the equation x=\frac{-7±17}{4} when ± is plus. Add -7 to 17.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-7±17}{4} when ± is minus. Subtract 17 from -7.

$$x=-\frac{24}{4}$$

Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=\frac{5}{2} x=-6$$

Steps for Completing the Square

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

$$2x^{2}+x-15=15-6x$$

Add 6x to both sides.

$$2x^{2}+x-15+6x=15$$

Combine x and 6x to get 7x.

$$2x^{2}+7x-15=15$$

Add 15 to both sides.

$$2x^{2}+7x=15+15$$

Add 15 and 15 to get 30.

$$2x^{2}+7x=30$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=\frac{30}{2}$$

Divide 30 by 2.

$$x^{2}+\frac{7}{2}x=15$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=15+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=15+\frac{49}{16}$$

Add 15 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{289}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{289}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{289}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{17}{4} x+\frac{7}{4}=-\frac{17}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{5}{2} x=-6$$