t^2+23t-14=0

Solution for t^2+23t-14=0 equation:

t^2+23t-14=0

a = 1; b = 23; c = -14;
Δ = b2-4ac
Δ = 232-4·1·(-14)
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-3\sqrt{65}}{2*1}=\frac{-23-3\sqrt{65}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+3\sqrt{65}}{2*1}=\frac{-23+3\sqrt{65}}{2}
$