Solve left(3x+4right)left(5x-7right)=left(2x+7right)^2+53

$$\left( 3x+4 \right) \left( 5x-7 \right) = { \left(2x+7 \right) }^{ 2 } +53$$

Solve for x

$$x = -\frac{26}{11} = -2\frac{4}{11} \approx -2.363636364$$

$$x=5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3x+4 by 5x-7 and combine like terms.

$$15x^{2}-x-28=\left(2x+7\right)^{2}+53$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$15x^{2}-x-28=4x^{2}+28x+49+53$$

Add 49 and 53 to get 102.

$$15x^{2}-x-28=4x^{2}+28x+102$$

Subtract 4x^{2} from both sides.

$$15x^{2}-x-28-4x^{2}=28x+102$$

Combine 15x^{2} and -4x^{2} to get 11x^{2}.

$$11x^{2}-x-28=28x+102$$

Subtract 28x from both sides.

$$11x^{2}-x-28-28x=102$$

Combine -x and -28x to get -29x.

$$11x^{2}-29x-28=102$$

Subtract 102 from both sides.

$$11x^{2}-29x-28-102=0$$

Subtract 102 from -28 to get -130.

$$11x^{2}-29x-130=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 11x^{2}+ax+bx-130. To find a and b, set up a system to be solved.

$$a+b=-29 ab=11\left(-130\right)=-1430$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1430.

$$1,-1430 2,-715 5,-286 10,-143 11,-130 13,-110 22,-65 26,-55$$

Calculate the sum for each pair.

$$1-1430=-1429 2-715=-713 5-286=-281 10-143=-133 11-130=-119 13-110=-97 22-65=-43 26-55=-29$$

The solution is the pair that gives sum -29.

$$a=-55 b=26$$

Rewrite 11x^{2}-29x-130 as \left(11x^{2}-55x\right)+\left(26x-130\right).

$$\left(11x^{2}-55x\right)+\left(26x-130\right)$$

Factor out 11x in the first and 26 in the second group.

$$11x\left(x-5\right)+26\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(11x+26\right)$$

To find equation solutions, solve x-5=0 and 11x+26=0.

$$x=5 x=-\frac{26}{11}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x+4 by 5x-7 and combine like terms.

$$15x^{2}-x-28=\left(2x+7\right)^{2}+53$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$15x^{2}-x-28=4x^{2}+28x+49+53$$

Add 49 and 53 to get 102.

$$15x^{2}-x-28=4x^{2}+28x+102$$

Subtract 4x^{2} from both sides.

$$15x^{2}-x-28-4x^{2}=28x+102$$

Combine 15x^{2} and -4x^{2} to get 11x^{2}.

$$11x^{2}-x-28=28x+102$$

Subtract 28x from both sides.

$$11x^{2}-x-28-28x=102$$

Combine -x and -28x to get -29x.

$$11x^{2}-29x-28=102$$

Subtract 102 from both sides.

$$11x^{2}-29x-28-102=0$$

Subtract 102 from -28 to get -130.

$$11x^{2}-29x-130=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, -29 for b, and -130 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 11\left(-130\right)}}{2\times 11}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 11\left(-130\right)}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-\left(-29\right)±\sqrt{841-44\left(-130\right)}}{2\times 11}$$

Multiply -44 times -130.

$$x=\frac{-\left(-29\right)±\sqrt{841+5720}}{2\times 11}$$

Add 841 to 5720.

$$x=\frac{-\left(-29\right)±\sqrt{6561}}{2\times 11}$$

Take the square root of 6561.

$$x=\frac{-\left(-29\right)±81}{2\times 11}$$

The opposite of -29 is 29.

$$x=\frac{29±81}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{29±81}{22}$$

Now solve the equation x=\frac{29±81}{22} when ± is plus. Add 29 to 81.

$$x=\frac{110}{22}$$

Divide 110 by 22.

$$x=5$$

Now solve the equation x=\frac{29±81}{22} when ± is minus. Subtract 81 from 29.

$$x=-\frac{52}{22}$$

Reduce the fraction \frac{-52}{22} to lowest terms by extracting and canceling out 2.

$$x=-\frac{26}{11}$$

The equation is now solved.

$$x=5 x=-\frac{26}{11}$$

Steps for Completing the Square

Use the distributive property to multiply 3x+4 by 5x-7 and combine like terms.

$$15x^{2}-x-28=\left(2x+7\right)^{2}+53$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$15x^{2}-x-28=4x^{2}+28x+49+53$$

Add 49 and 53 to get 102.

$$15x^{2}-x-28=4x^{2}+28x+102$$

Subtract 4x^{2} from both sides.

$$15x^{2}-x-28-4x^{2}=28x+102$$

Combine 15x^{2} and -4x^{2} to get 11x^{2}.

$$11x^{2}-x-28=28x+102$$

Subtract 28x from both sides.

$$11x^{2}-x-28-28x=102$$

Combine -x and -28x to get -29x.

$$11x^{2}-29x-28=102$$

Add 28 to both sides.

$$11x^{2}-29x=102+28$$

Add 102 and 28 to get 130.

$$11x^{2}-29x=130$$

Divide both sides by 11.

$$\frac{11x^{2}-29x}{11}=\frac{130}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$x^{2}-\frac{29}{11}x=\frac{130}{11}$$

Divide -\frac{29}{11}, the coefficient of the x term, by 2 to get -\frac{29}{22}. Then add the square of -\frac{29}{22} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{29}{11}x+\left(-\frac{29}{22}\right)^{2}=\frac{130}{11}+\left(-\frac{29}{22}\right)^{2}$$

Square -\frac{29}{22} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{29}{11}x+\frac{841}{484}=\frac{130}{11}+\frac{841}{484}$$

Add \frac{130}{11} to \frac{841}{484} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{29}{11}x+\frac{841}{484}=\frac{6561}{484}$$

Factor x^{2}-\frac{29}{11}x+\frac{841}{484}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{29}{22}\right)^{2}=\frac{6561}{484}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{29}{22}\right)^{2}}=\sqrt{\frac{6561}{484}}$$

Simplify.

$$x-\frac{29}{22}=\frac{81}{22} x-\frac{29}{22}=-\frac{81}{22}$$

Add \frac{29}{22} to both sides of the equation.

$$x=5 x=-\frac{26}{11}$$