(7t-4t^2)+15*(3t+3)=0

Solution for (7t-4t^2)+15*(3t+3)=0 equation:

(7t-4t^2)+15(3t+3)=0

We multiply parentheses

(7t-4t^2)+45t+45=0

We get rid of parentheses

-4t^2+7t+45t+45=0

We add all the numbers together, and all the variables

-4t^2+52t+45=0

a = -4; b = 52; c = +45;
Δ = b2-4ac
Δ = 522-4·(-4)·45
Δ = 3424
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:

$\sqrt{\Delta}=\sqrt{3424}=\sqrt{16*214}=\sqrt{16}*\sqrt{214}=4\sqrt{214}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{214}}{2*-4}=\frac{-52-4\sqrt{214}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{214}}{2*-4}=\frac{-52+4\sqrt{214}}{-8}
$