3t2+25t=18

Solution for 3t2+25t=18 equation:

3t^2+25t=18

We move all terms to the left:

3t^2+25t-(18)=0

a = 3; b = 25; c = -18;
Δ = b2-4ac
Δ = 252-4·3·(-18)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-29}{2*3}=\frac{-54}{6}
=-9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+29}{2*3}=\frac{4}{6}
=2/3
$