Solve 22x=8x^2+15

$$22 x = 8 x ^ { 2 } + 15$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Subtract 8x^{2} from both sides.

$$22x-8x^{2}=15$$

Subtract 15 from both sides.

$$22x-8x^{2}-15=0$$

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

$$-8x^{2}+22x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=22 ab=-8\left(-15\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 22.

$$a=12 b=10$$

Rewrite -8x^{2}+22x-15 as \left(-8x^{2}+12x\right)+\left(10x-15\right).

$$\left(-8x^{2}+12x\right)+\left(10x-15\right)$$

Factor out -4x in the first and 5 in the second group.

$$-4x\left(2x-3\right)+5\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-4x+5\right)$$

To find equation solutions, solve 2x-3=0 and -4x+5=0.

$$x=\frac{3}{2} x=\frac{5}{4}$$

Steps Using the Quadratic Formula

Subtract 8x^{2} from both sides.

$$22x-8x^{2}=15$$

Subtract 15 from both sides.

$$22x-8x^{2}-15=0$$

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.

$$-8x^{2}+22x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 22 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-22±\sqrt{22^{2}-4\left(-8\right)\left(-15\right)}}{2\left(-8\right)}$$

Square 22.

$$x=\frac{-22±\sqrt{484-4\left(-8\right)\left(-15\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-22±\sqrt{484+32\left(-15\right)}}{2\left(-8\right)}$$

Multiply 32 times -15.

$$x=\frac{-22±\sqrt{484-480}}{2\left(-8\right)}$$

Add 484 to -480.

$$x=\frac{-22±\sqrt{4}}{2\left(-8\right)}$$

Take the square root of 4.

$$x=\frac{-22±2}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-22±2}{-16}$$

Now solve the equation x=\frac{-22±2}{-16} when ± is plus. Add -22 to 2.

$$x=-\frac{20}{-16}$$

Reduce the fraction \frac{-20}{-16} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{-22±2}{-16} when ± is minus. Subtract 2 from -22.

$$x=-\frac{24}{-16}$$

Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=\frac{5}{4} x=\frac{3}{2}$$

Steps for Completing the Square

Subtract 8x^{2} from both sides.

$$22x-8x^{2}=15$$

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.

$$-8x^{2}+22x=15$$

Divide both sides by -8.

$$\frac{-8x^{2}+22x}{-8}=\frac{15}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{22}{-8}x=\frac{15}{-8}$$

Reduce the fraction \frac{22}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{4}x=\frac{15}{-8}$$

Divide 15 by -8.

$$x^{2}-\frac{11}{4}x=-\frac{15}{8}$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=-\frac{15}{8}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=-\frac{15}{8}+\frac{121}{64}$$

Add -\frac{15}{8} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{1}{8} x-\frac{11}{8}=-\frac{1}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{5}{4}$$