0.75t^2+2t-8=0

Solution for 0.75t^2+2t-8=0 equation:

0.75t^2+2t-8=0

a = 0.75; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·0.75·(-8)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*0.75}=\frac{-2-2\sqrt{7}}{1.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*0.75}=\frac{-2+2\sqrt{7}}{1.5}
$