2t^2-3t=15

Solution for 2t^2-3t=15 equation:

2t^2-3t=15

We move all terms to the left:

2t^2-3t-(15)=0

a = 2; b = -3; c = -15;
Δ = b2-4ac
Δ = -32-4·2·(-15)
Δ = 129
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{-(-3)-\sqrt{129}}{2*2}=\frac{3-\sqrt{129}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*2}=\frac{3+\sqrt{129}}{4}
$