Solve (4x-3)-5=(2x+5)(6x-7)

$$( 4 x - 3 ) - 5 = ( 2 x + 5 ) ( 6 x - 7 )$$

Solve for x

$$x = \frac{\sqrt{10} - 1}{2} \approx 1.08113883$$

$$x=\frac{-\sqrt{10}-1}{2}\approx -2.08113883$$

Steps Using the Quadratic Formula

Subtract 5 from -3 to get -8.

$$4x-8=\left(2x+5\right)\left(6x-7\right)$$

Use the distributive property to multiply 2x+5 by 6x-7 and combine like terms.

$$4x-8=12x^{2}+16x-35$$

Subtract 12x^{2} from both sides.

$$4x-8-12x^{2}=16x-35$$

Subtract 16x from both sides.

$$4x-8-12x^{2}-16x=-35$$

Combine 4x and -16x to get -12x.

$$-12x-8-12x^{2}=-35$$

Add 35 to both sides.

$$-12x-8-12x^{2}+35=0$$

Add -8 and 35 to get 27.

$$-12x+27-12x^{2}=0$$

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.

$$-12x^{2}-12x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -12 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-12\right)\times 27}}{2\left(-12\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-12\right)\times 27}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-12\right)±\sqrt{144+48\times 27}}{2\left(-12\right)}$$

Multiply 48 times 27.

$$x=\frac{-\left(-12\right)±\sqrt{144+1296}}{2\left(-12\right)}$$

Add 144 to 1296.

$$x=\frac{-\left(-12\right)±\sqrt{1440}}{2\left(-12\right)}$$

Take the square root of 1440.

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

The opposite of -12 is 12.

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

Multiply 2 times -12.

$$x=\frac{12±12\sqrt{10}}{-24}$$

Now solve the equation x=\frac{12±12\sqrt{10}}{-24} when ± is plus. Add 12 to 12\sqrt{10}.

$$x=\frac{12\sqrt{10}+12}{-24}$$

Divide 12+12\sqrt{10} by -24.

$$x=\frac{-\sqrt{10}-1}{2}$$

Now solve the equation x=\frac{12±12\sqrt{10}}{-24} when ± is minus. Subtract 12\sqrt{10} from 12.

$$x=\frac{12-12\sqrt{10}}{-24}$$

Divide 12-12\sqrt{10} by -24.

$$x=\frac{\sqrt{10}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{10}-1}{2} x=\frac{\sqrt{10}-1}{2}$$

Steps for Completing the Square

Subtract 5 from -3 to get -8.

$$4x-8=\left(2x+5\right)\left(6x-7\right)$$

Use the distributive property to multiply 2x+5 by 6x-7 and combine like terms.

$$4x-8=12x^{2}+16x-35$$

Subtract 12x^{2} from both sides.

$$4x-8-12x^{2}=16x-35$$

Subtract 16x from both sides.

$$4x-8-12x^{2}-16x=-35$$

Combine 4x and -16x to get -12x.

$$-12x-8-12x^{2}=-35$$

Add 8 to both sides.

$$-12x-12x^{2}=-35+8$$

Add -35 and 8 to get -27.

$$-12x-12x^{2}=-27$$

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.

$$-12x^{2}-12x=-27$$

Divide both sides by -12.

$$\frac{-12x^{2}-12x}{-12}=-\frac{27}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{12}{-12}\right)x=-\frac{27}{-12}$$

Divide -12 by -12.

$$x^{2}+x=-\frac{27}{-12}$$

Reduce the fraction \frac{-27}{-12} to lowest terms by extracting and canceling out 3.

$$x^{2}+x=\frac{9}{4}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{9}{4}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{9+1}{4}$$

Add \frac{9}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{5}{2}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{2}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{10}}{2} x+\frac{1}{2}=-\frac{\sqrt{10}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{10}-1}{2} x=\frac{-\sqrt{10}-1}{2}$$