Solve Using the Quadratic Formula 10b^2=27b-18

$10 b^{2} = 27 b - 18$

Step 1

Move all the expressions to the left side of the equation.

Subtract $27 b$ from both sides of the equation.

$10 b^{2} - 27 b = - 18$

Add $18$ to both sides of the equation.

$10 b^{2} - 27 b + 18 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 10$ , $b = - 27$ , and $c = 18$ into the quadratic formula and solve for $b$ .

$\frac{27 \pm \sqrt{{(- 27)}^{2} - 4 \cdot (10 \cdot 18)}}{2 \cdot 10}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 27$ to the power of $2$ .

$b = \frac{27 \pm \sqrt{729 - 4 \cdot 10 \cdot 18}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot 18$ .

Multiply $- 4$ by $10$ .

$b = \frac{27 \pm \sqrt{729 - 40 \cdot 18}}{2 \cdot 10}$

Multiply $- 40$ by $18$ .

$b = \frac{27 \pm \sqrt{729 - 720}}{2 \cdot 10}$

Subtract $720$ from $729$ .

$b = \frac{27 \pm \sqrt{9}}{2 \cdot 10}$

Rewrite $9$ as $3^{2}$ .

$b = \frac{27 \pm \sqrt{3^{2}}}{2 \cdot 10}$

Pull terms out from under the radical, assuming positive real numbers.

$b = \frac{27 \pm 3}{2 \cdot 10}$

Multiply $2$ by $10$ .

$b = \frac{27 \pm 3}{20}$

Step 5

The final answer is the combination of both solutions.

$b = \frac{3}{2}, \frac{6}{5}$