Solve 3x(x+5)/5-left(2x+1right)^2/4=(x-4)(x+4)/2

$$\frac{ 3x(x+5) }{ 5 } - \frac{ { \left(2x+1 \right) }^{ 2 } }{ 4 } = \frac{ (x-4)(x+4) }{ 2 }$$

Solve for x

$$x=\frac{\sqrt{3190}}{18}+\frac{10}{9}\approx 4.24889361$$

$$x=-\frac{\sqrt{3190}}{18}+\frac{10}{9}\approx -2.026671388$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 20, the least common multiple of 5,4,2.

$$4\times 3x\left(x+5\right)-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

Multiply 4 and 3 to get 12.

$$12x\left(x+5\right)-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply 12x by x+5.

$$12x^{2}+60x-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

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

$$12x^{2}+60x-5\left(4x^{2}+4x+1\right)=10\left(x-4\right)\left(x+4\right)$$

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

$$12x^{2}+60x-20x^{2}-20x-5=10\left(x-4\right)\left(x+4\right)$$

Combine 12x^{2} and -20x^{2} to get -8x^{2}.

$$-8x^{2}+60x-20x-5=10\left(x-4\right)\left(x+4\right)$$

Combine 60x and -20x to get 40x.

$$-8x^{2}+40x-5=10\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply 10 by x-4.

$$-8x^{2}+40x-5=\left(10x-40\right)\left(x+4\right)$$

Use the distributive property to multiply 10x-40 by x+4 and combine like terms.

$$-8x^{2}+40x-5=10x^{2}-160$$

Subtract 10x^{2} from both sides.

$$-8x^{2}+40x-5-10x^{2}=-160$$

Combine -8x^{2} and -10x^{2} to get -18x^{2}.

$$-18x^{2}+40x-5=-160$$

Add 160 to both sides.

$$-18x^{2}+40x-5+160=0$$

Add -5 and 160 to get 155.

$$-18x^{2}+40x+155=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 40 for b, and 155 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-18\right)\times 155}}{2\left(-18\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-18\right)\times 155}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-40±\sqrt{1600+72\times 155}}{2\left(-18\right)}$$

Multiply 72 times 155.

$$x=\frac{-40±\sqrt{1600+11160}}{2\left(-18\right)}$$

Add 1600 to 11160.

$$x=\frac{-40±\sqrt{12760}}{2\left(-18\right)}$$

Take the square root of 12760.

$$x=\frac{-40±2\sqrt{3190}}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{-40±2\sqrt{3190}}{-36}$$

Now solve the equation x=\frac{-40±2\sqrt{3190}}{-36} when ± is plus. Add -40 to 2\sqrt{3190}.

$$x=\frac{2\sqrt{3190}-40}{-36}$$

Divide -40+2\sqrt{3190} by -36.

$$x=-\frac{\sqrt{3190}}{18}+\frac{10}{9}$$

Now solve the equation x=\frac{-40±2\sqrt{3190}}{-36} when ± is minus. Subtract 2\sqrt{3190} from -40.

$$x=\frac{-2\sqrt{3190}-40}{-36}$$

Divide -40-2\sqrt{3190} by -36.

$$x=\frac{\sqrt{3190}}{18}+\frac{10}{9}$$

The equation is now solved.

$$x=-\frac{\sqrt{3190}}{18}+\frac{10}{9} x=\frac{\sqrt{3190}}{18}+\frac{10}{9}$$

Steps for Completing the Square

Multiply both sides of the equation by 20, the least common multiple of 5,4,2.

$$4\times 3x\left(x+5\right)-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

Multiply 4 and 3 to get 12.

$$12x\left(x+5\right)-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply 12x by x+5.

$$12x^{2}+60x-5\left(2x+1\right)^{2}=10\left(x-4\right)\left(x+4\right)$$

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

$$12x^{2}+60x-5\left(4x^{2}+4x+1\right)=10\left(x-4\right)\left(x+4\right)$$

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

$$12x^{2}+60x-20x^{2}-20x-5=10\left(x-4\right)\left(x+4\right)$$

Combine 12x^{2} and -20x^{2} to get -8x^{2}.

$$-8x^{2}+60x-20x-5=10\left(x-4\right)\left(x+4\right)$$

Combine 60x and -20x to get 40x.

$$-8x^{2}+40x-5=10\left(x-4\right)\left(x+4\right)$$

Use the distributive property to multiply 10 by x-4.

$$-8x^{2}+40x-5=\left(10x-40\right)\left(x+4\right)$$

Use the distributive property to multiply 10x-40 by x+4 and combine like terms.

$$-8x^{2}+40x-5=10x^{2}-160$$

Subtract 10x^{2} from both sides.

$$-8x^{2}+40x-5-10x^{2}=-160$$

Combine -8x^{2} and -10x^{2} to get -18x^{2}.

$$-18x^{2}+40x-5=-160$$

Add 5 to both sides.

$$-18x^{2}+40x=-160+5$$

Add -160 and 5 to get -155.

$$-18x^{2}+40x=-155$$

Divide both sides by -18.

$$\frac{-18x^{2}+40x}{-18}=-\frac{155}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\frac{40}{-18}x=-\frac{155}{-18}$$

Reduce the fraction \frac{40}{-18} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{20}{9}x=-\frac{155}{-18}$$

Divide -155 by -18.

$$x^{2}-\frac{20}{9}x=\frac{155}{18}$$

Divide -\frac{20}{9}, the coefficient of the x term, by 2 to get -\frac{10}{9}. Then add the square of -\frac{10}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{9}x+\left(-\frac{10}{9}\right)^{2}=\frac{155}{18}+\left(-\frac{10}{9}\right)^{2}$$

Square -\frac{10}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{9}x+\frac{100}{81}=\frac{155}{18}+\frac{100}{81}$$

Add \frac{155}{18} to \frac{100}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{9}x+\frac{100}{81}=\frac{1595}{162}$$

Factor x^{2}-\frac{20}{9}x+\frac{100}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{9}\right)^{2}=\frac{1595}{162}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{9}\right)^{2}}=\sqrt{\frac{1595}{162}}$$

Simplify.

$$x-\frac{10}{9}=\frac{\sqrt{3190}}{18} x-\frac{10}{9}=-\frac{\sqrt{3190}}{18}$$

Add \frac{10}{9} to both sides of the equation.

$$x=\frac{\sqrt{3190}}{18}+\frac{10}{9} x=-\frac{\sqrt{3190}}{18}+\frac{10}{9}$$