3t^2+10t-57=0

Solution for 3t^2+10t-57=0 equation:

3t^2+10t-57=0

a = 3; b = 10; c = -57;
Δ = b2-4ac
Δ = 102-4·3·(-57)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-28}{2*3}=\frac{-38}{6}
=-6+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+28}{2*3}=\frac{18}{6}
=3
$