Solve Using the Quadratic Formula 10g^2+10=29g

$10 g^{2} + 10 = 29 g$

Step 1

Subtract $29 g$ from both sides of the equation.

$10 g^{2} + 10 - 29 g = 0$

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 $a = 10$ , $b = - 29$ , and $c = 10$ into the quadratic formula and solve for $g$ .

$\frac{29 \pm \sqrt{{(- 29)}^{2} - 4 \cdot (10 \cdot 10)}}{2 \cdot 10}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 29$ to the power of $2$ .

$g = \frac{29 \pm \sqrt{841 - 4 \cdot 10 \cdot 10}}{2 \cdot 10}$

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

Multiply $- 4$ by $10$ .

$g = \frac{29 \pm \sqrt{841 - 40 \cdot 10}}{2 \cdot 10}$

Multiply $- 40$ by $10$ .

$g = \frac{29 \pm \sqrt{841 - 400}}{2 \cdot 10}$

Subtract $400$ from $841$ .

$g = \frac{29 \pm \sqrt{441}}{2 \cdot 10}$

Rewrite $441$ as $21^{2}$ .

$g = \frac{29 \pm \sqrt{21^{2}}}{2 \cdot 10}$

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

$g = \frac{29 \pm 21}{2 \cdot 10}$

Multiply $2$ by $10$ .

$g = \frac{29 \pm 21}{20}$

Step 5

The final answer is the combination of both solutions.

$g = \frac{5}{2}, \frac{2}{5}$