4t^2+4t-30=0

Solution for 4t^2+4t-30=0 equation:

4t^2+4t-30=0

a = 4; b = 4; c = -30;
Δ = b2-4ac
Δ = 42-4·4·(-30)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*4}=\frac{-4-4\sqrt{31}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*4}=\frac{-4+4\sqrt{31}}{8}
$