t^2=9t+14=0

Solution for t^2=9t+14=0 equation:

t^2=9t+14=0

We move all terms to the left:

t^2-(9t+14)=0

We get rid of parentheses

t^2-9t-14=0

a = 1; b = -9; c = -14;
Δ = b2-4ac
Δ = -92-4·1·(-14)
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{137}}{2*1}=\frac{9-\sqrt{137}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{137}}{2*1}=\frac{9+\sqrt{137}}{2}
$