Solve Using the Quadratic Formula 10x^2+21x-10=0

$10 x^{2} + 21 x - 10 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 10$ , $b = 21$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 21 \pm \sqrt{21^{2} - 4 \cdot (10 \cdot - 10)}}{2 \cdot 10}$

Step 3

Simplify.

Simplify the numerator.

Raise $21$ to the power of $2$ .

$x = \frac{- 21 \pm \sqrt{441 - 4 \cdot 10 \cdot - 10}}{2 \cdot 10}$

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

Multiply $- 4$ by $10$ .

$x = \frac{- 21 \pm \sqrt{441 - 40 \cdot - 10}}{2 \cdot 10}$

Multiply $- 40$ by $- 10$ .

$x = \frac{- 21 \pm \sqrt{441 + 400}}{2 \cdot 10}$

Add $441$ and $400$ .

$x = \frac{- 21 \pm \sqrt{841}}{2 \cdot 10}$

Rewrite $841$ as $29^{2}$ .

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

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

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

Multiply $2$ by $10$ .

$x = \frac{- 21 \pm 29}{20}$

Step 4

The final answer is the combination of both solutions.

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