-3t(t-5)+2(7+2t)=36

Solution for -3t(t-5)+2(7+2t)=36 equation:

-3t(t-5)+2(7+2t)=36

We move all terms to the left:

-3t(t-5)+2(7+2t)-(36)=0

We add all the numbers together, and all the variables

-3t(t-5)+2(2t+7)-36=0

We multiply parentheses

-3t^2+15t+4t+14-36=0

We add all the numbers together, and all the variables

-3t^2+19t-22=0

a = -3; b = 19; c = -22;
Δ = b2-4ac
Δ = 192-4·(-3)·(-22)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{97}}{2*-3}=\frac{-19-\sqrt{97}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{97}}{2*-3}=\frac{-19+\sqrt{97}}{-6}
$