10t^2+2t-48=0

Solution for 10t^2+2t-48=0 equation:

10t^2+2t-48=0

a = 10; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·10·(-48)
Δ = 1924
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:

$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{481}}{2*10}=\frac{-2-2\sqrt{481}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{481}}{2*10}=\frac{-2+2\sqrt{481}}{20}
$