t^2+12t-15=0

Solution for t^2+12t-15=0 equation:

t^2+12t-15=0

a = 1; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·1·(-15)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{51}}{2*1}=\frac{-12-2\sqrt{51}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{51}}{2*1}=\frac{-12+2\sqrt{51}}{2}
$