-16t^2+23t+4=8

Solution for -16t^2+23t+4=8 equation:

-16t^2+23t+4=8

We move all terms to the left:

-16t^2+23t+4-(8)=0

We add all the numbers together, and all the variables

-16t^2+23t-4=0

a = -16; b = 23; c = -4;
Δ = b2-4ac
Δ = 232-4·(-16)·(-4)
Δ = 273
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{273}}{2*-16}=\frac{-23-\sqrt{273}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{273}}{2*-16}=\frac{-23+\sqrt{273}}{-32}
$