4t^2-28t-48=0

Solution for 4t^2-28t-48=0 equation:

4t^2-28t-48=0

a = 4; b = -28; c = -48;
Δ = b2-4ac
Δ = -282-4·4·(-48)
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{97}}{2*4}=\frac{28-4\sqrt{97}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{97}}{2*4}=\frac{28+4\sqrt{97}}{8}
$