16t^2+23t-3=11

Solution for 16t^2+23t-3=11 equation:

16t^2+23t-3=11

We move all terms to the left:

16t^2+23t-3-(11)=0

We add all the numbers together, and all the variables

16t^2+23t-14=0

a = 16; b = 23; c = -14;
Δ = b2-4ac
Δ = 232-4·16·(-14)
Δ = 1425
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:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-5\sqrt{57}}{2*16}=\frac{-23-5\sqrt{57}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+5\sqrt{57}}{2*16}=\frac{-23+5\sqrt{57}}{32}
$