23=2t^2+7t-8

Solution for 23=2t^2+7t-8 equation:

23=2t^2+7t-8

We move all terms to the left:

23-(2t^2+7t-8)=0

We get rid of parentheses

-2t^2-7t+8+23=0

We add all the numbers together, and all the variables

-2t^2-7t+31=0

a = -2; b = -7; c = +31;
Δ = b2-4ac
Δ = -72-4·(-2)·31
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3\sqrt{33}}{2*-2}=\frac{7-3\sqrt{33}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3\sqrt{33}}{2*-2}=\frac{7+3\sqrt{33}}{-4}
$