23=7t+16t^2

Solution for 23=7t+16t^2 equation:

23=7t+16t^2

We move all terms to the left:

23-(7t+16t^2)=0

We get rid of parentheses

-16t^2-7t+23=0

a = -16; b = -7; c = +23;
Δ = b2-4ac
Δ = -72-4·(-16)·23
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-39}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+39}{2*-16}=\frac{46}{-32}
=-1+7/16
$