Solve for t at^2+bt=-c

$a t^{2} + b t = - c$

Move $c$ to the left side of the equation by adding it to both sides.

$a t^{2} + b t + c = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = a$ , $b = b$ , and $c = c$ into the quadratic formula and solve for $t$ .

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

Remove parentheses.

$t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

The final answer is the combination of both solutions.

$t = - \frac{b - \sqrt{b^{2} - 4 a c}}{2 a}$

$t = - \frac{b + \sqrt{b^{2} - 4 a c}}{2 a}$

Solve for t at^2+bt=-c

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