16t^2+2t-52.5=0

Solution for 16t^2+2t-52.5=0 equation:

16t^2+2t-52.5=0

a = 16; b = 2; c = -52.5;
Δ = b2-4ac
Δ = 22-4·16·(-52.5)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-58}{2*16}=\frac{-60}{32}
=-1+7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+58}{2*16}=\frac{56}{32}
=1+3/4
$