Solve 2x^2+5+20x=20-8x^2+x

$$2 x ^ { 2 } + 5 + 20 x = 20 - 8 x ^ { 2 } + x$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Subtract 20 from both sides.

$$2x^{2}+5+20x-20=-8x^{2}+x$$

Subtract 20 from 5 to get -15.

$$2x^{2}-15+20x=-8x^{2}+x$$

Add 8x^{2} to both sides.

$$2x^{2}-15+20x+8x^{2}=x$$

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

$$10x^{2}-15+20x=x$$

Subtract x from both sides.

$$10x^{2}-15+20x-x=0$$

Combine 20x and -x to get 19x.

$$10x^{2}-15+19x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$10x^{2}+19x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=19 ab=10\left(-15\right)=-150$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -150.

$$-1,150 -2,75 -3,50 -5,30 -6,25 -10,15$$

Calculate the sum for each pair.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

The solution is the pair that gives sum 19.

$$a=-6 b=25$$

Rewrite 10x^{2}+19x-15 as \left(10x^{2}-6x\right)+\left(25x-15\right).

$$\left(10x^{2}-6x\right)+\left(25x-15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(5x-3\right)+5\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(2x+5\right)$$

To find equation solutions, solve 5x-3=0 and 2x+5=0.

$$x=\frac{3}{5} x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 20 from both sides.

$$2x^{2}+5+20x-20=-8x^{2}+x$$

Subtract 20 from 5 to get -15.

$$2x^{2}-15+20x=-8x^{2}+x$$

Add 8x^{2} to both sides.

$$2x^{2}-15+20x+8x^{2}=x$$

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

$$10x^{2}-15+20x=x$$

Subtract x from both sides.

$$10x^{2}-15+20x-x=0$$

Combine 20x and -x to get 19x.

$$10x^{2}-15+19x=0$$

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.

$$10x^{2}+19x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 19 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-19±\sqrt{19^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-19±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

Multiply -40 times -15.

$$x=\frac{-19±\sqrt{361+600}}{2\times 10}$$

Add 361 to 600.

$$x=\frac{-19±\sqrt{961}}{2\times 10}$$

Take the square root of 961.

$$x=\frac{-19±31}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-19±31}{20}$$

Now solve the equation x=\frac{-19±31}{20} when ± is plus. Add -19 to 31.

$$x=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-19±31}{20} when ± is minus. Subtract 31 from -19.

$$x=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=\frac{3}{5} x=-\frac{5}{2}$$

Steps for Completing the Square

Add 8x^{2} to both sides.

$$2x^{2}+5+20x+8x^{2}=20+x$$

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

$$10x^{2}+5+20x=20+x$$

Subtract x from both sides.

$$10x^{2}+5+20x-x=20$$

Combine 20x and -x to get 19x.

$$10x^{2}+5+19x=20$$

Subtract 5 from both sides.

$$10x^{2}+19x=20-5$$

Subtract 5 from 20 to get 15.

$$10x^{2}+19x=15$$

Divide both sides by 10.

$$\frac{10x^{2}+19x}{10}=\frac{15}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{19}{10}x=\frac{15}{10}$$

Reduce the fraction \frac{15}{10} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{19}{10}x=\frac{3}{2}$$

Divide \frac{19}{10}, the coefficient of the x term, by 2 to get \frac{19}{20}. Then add the square of \frac{19}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{10}x+\left(\frac{19}{20}\right)^{2}=\frac{3}{2}+\left(\frac{19}{20}\right)^{2}$$

Square \frac{19}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{10}x+\frac{361}{400}=\frac{3}{2}+\frac{361}{400}$$

Add \frac{3}{2} to \frac{361}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{10}x+\frac{361}{400}=\frac{961}{400}$$

Factor x^{2}+\frac{19}{10}x+\frac{361}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{20}\right)^{2}=\frac{961}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{20}\right)^{2}}=\sqrt{\frac{961}{400}}$$

Simplify.

$$x+\frac{19}{20}=\frac{31}{20} x+\frac{19}{20}=-\frac{31}{20}$$

Subtract \frac{19}{20} from both sides of the equation.

$$x=\frac{3}{5} x=-\frac{5}{2}$$