Solve 3(12x+13)=2(x+5)(2x-5)

$$3 ( 12 x + 13 ) = 2 ( x + 5 ) ( 2 x - 5 )$$

Solve for x

$$x = \frac{5 \sqrt{21} + 13}{4} \approx 8.978219619$$

$$x=\frac{13-5\sqrt{21}}{4}\approx -2.478219619$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 12x+13.

$$36x+39=2\left(x+5\right)\left(2x-5\right)$$

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

$$36x+39=\left(2x+10\right)\left(2x-5\right)$$

Use the distributive property to multiply 2x+10 by 2x-5 and combine like terms.

$$36x+39=4x^{2}+10x-50$$

Subtract 4x^{2} from both sides.

$$36x+39-4x^{2}=10x-50$$

Subtract 10x from both sides.

$$36x+39-4x^{2}-10x=-50$$

Combine 36x and -10x to get 26x.

$$26x+39-4x^{2}=-50$$

Add 50 to both sides.

$$26x+39-4x^{2}+50=0$$

Add 39 and 50 to get 89.

$$26x+89-4x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4x^{2}+26x+89=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 26 for b, and 89 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-4\right)\times 89}}{2\left(-4\right)}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-4\right)\times 89}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-26±\sqrt{676+16\times 89}}{2\left(-4\right)}$$

Multiply 16 times 89.

$$x=\frac{-26±\sqrt{676+1424}}{2\left(-4\right)}$$

Add 676 to 1424.

$$x=\frac{-26±\sqrt{2100}}{2\left(-4\right)}$$

Take the square root of 2100.

$$x=\frac{-26±10\sqrt{21}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-26±10\sqrt{21}}{-8}$$

Now solve the equation x=\frac{-26±10\sqrt{21}}{-8} when ± is plus. Add -26 to 10\sqrt{21}.

$$x=\frac{10\sqrt{21}-26}{-8}$$

Divide -26+10\sqrt{21} by -8.

$$x=\frac{13-5\sqrt{21}}{4}$$

Now solve the equation x=\frac{-26±10\sqrt{21}}{-8} when ± is minus. Subtract 10\sqrt{21} from -26.

$$x=\frac{-10\sqrt{21}-26}{-8}$$

Divide -26-10\sqrt{21} by -8.

$$x=\frac{5\sqrt{21}+13}{4}$$

The equation is now solved.

$$x=\frac{13-5\sqrt{21}}{4} x=\frac{5\sqrt{21}+13}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by 12x+13.

$$36x+39=2\left(x+5\right)\left(2x-5\right)$$

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

$$36x+39=\left(2x+10\right)\left(2x-5\right)$$

Use the distributive property to multiply 2x+10 by 2x-5 and combine like terms.

$$36x+39=4x^{2}+10x-50$$

Subtract 4x^{2} from both sides.

$$36x+39-4x^{2}=10x-50$$

Subtract 10x from both sides.

$$36x+39-4x^{2}-10x=-50$$

Combine 36x and -10x to get 26x.

$$26x+39-4x^{2}=-50$$

Subtract 39 from both sides.

$$26x-4x^{2}=-50-39$$

Subtract 39 from -50 to get -89.

$$26x-4x^{2}=-89$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+26x=-89$$

Divide both sides by -4.

$$\frac{-4x^{2}+26x}{-4}=-\frac{89}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{26}{-4}x=-\frac{89}{-4}$$

Reduce the fraction \frac{26}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{13}{2}x=-\frac{89}{-4}$$

Divide -89 by -4.

$$x^{2}-\frac{13}{2}x=\frac{89}{4}$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=\frac{89}{4}+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{89}{4}+\frac{169}{16}$$

Add \frac{89}{4} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{525}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{525}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{525}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{5\sqrt{21}}{4} x-\frac{13}{4}=-\frac{5\sqrt{21}}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{5\sqrt{21}+13}{4} x=\frac{13-5\sqrt{21}}{4}$$