(7t-9t^2)+13(5t+2)=

Solution for (7t-9t^2)+13(5t+2)= equation:

(7t-9t^2)+13(5t+2)=

We move all terms to the left:

(7t-9t^2)+13(5t+2)-()=0

We add all the numbers together, and all the variables

(7t-9t^2)+13(5t+2)=0

We multiply parentheses

(7t-9t^2)+65t+26=0

We get rid of parentheses

-9t^2+7t+65t+26=0

We add all the numbers together, and all the variables

-9t^2+72t+26=0

a = -9; b = 72; c = +26;
Δ = b2-4ac
Δ = 722-4·(-9)·26
Δ = 6120
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:

$\sqrt{\Delta}=\sqrt{6120}=\sqrt{36*170}=\sqrt{36}*\sqrt{170}=6\sqrt{170}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-6\sqrt{170}}{2*-9}=\frac{-72-6\sqrt{170}}{-18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+6\sqrt{170}}{2*-9}=\frac{-72+6\sqrt{170}}{-18}
$