Solve quad5(2x+1)^2/4x+5=5x-1

$$\quad \frac { 5 ( 2 x + 1 ) ^ { 2 } } { 4 x + 5 } = 5 x - 1$$

Solve for x

$$x=10$$

Steps for Solving Linear Equation

Variable x cannot be equal to -\frac{5}{4} since division by zero is not defined. Multiply both sides of the equation by 4x+5.

$$5\left(2x+1\right)^{2}=5x\left(4x+5\right)+\left(4x+5\right)\left(-1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$5\left(4x^{2}+4x+1\right)=5x\left(4x+5\right)+\left(4x+5\right)\left(-1\right)$$

Use the distributive property to multiply 5 by 4x^{2}+4x+1.

$$20x^{2}+20x+5=5x\left(4x+5\right)+\left(4x+5\right)\left(-1\right)$$

Use the distributive property to multiply 5x by 4x+5.

$$20x^{2}+20x+5=20x^{2}+25x+\left(4x+5\right)\left(-1\right)$$

Use the distributive property to multiply 4x+5 by -1.

$$20x^{2}+20x+5=20x^{2}+25x-4x-5$$

Combine 25x and -4x to get 21x.

$$20x^{2}+20x+5=20x^{2}+21x-5$$

Subtract 20x^{2} from both sides.

$$20x^{2}+20x+5-20x^{2}=21x-5$$

Combine 20x^{2} and -20x^{2} to get 0.

$$20x+5=21x-5$$

Subtract 21x from both sides.

$$20x+5-21x=-5$$

Combine 20x and -21x to get -x.

$$-x+5=-5$$

Subtract 5 from both sides.

$$-x=-5-5$$

Subtract 5 from -5 to get -10.

$$-x=-10$$

Multiply both sides by -1.

$$x=10$$