2t^2=25t+12

Solution for 2t^2=25t+12 equation:

2t^2=25t+12

We move all terms to the left:

2t^2-(25t+12)=0

We get rid of parentheses

2t^2-25t-12=0

a = 2; b = -25; c = -12;
Δ = b2-4ac
Δ = -252-4·2·(-12)
Δ = 721
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{-(-25)-\sqrt{721}}{2*2}=\frac{25-\sqrt{721}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{721}}{2*2}=\frac{25+\sqrt{721}}{4}
$