15=7t+4t^2

Solution for 15=7t+4t^2 equation:

15=7t+4t^2

We move all terms to the left:

15-(7t+4t^2)=0

We get rid of parentheses

-4t^2-7t+15=0

a = -4; b = -7; c = +15;
Δ = b2-4ac
Δ = -72-4·(-4)·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*-4}=\frac{-10}{-8}
=1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*-4}=\frac{24}{-8}
=-3
$