2t+3t+20=2t^2-6t

Solution for 2t+3t+20=2t^2-6t equation:

2t+3t+20=2t^2-6t

We move all terms to the left:

2t+3t+20-(2t^2-6t)=0

We add all the numbers together, and all the variables

5t-(2t^2-6t)+20=0

We get rid of parentheses

-2t^2+5t+6t+20=0

We add all the numbers together, and all the variables

-2t^2+11t+20=0

a = -2; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·(-2)·20
Δ = 281
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{-(11)-\sqrt{281}}{2*-2}=\frac{-11-\sqrt{281}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{281}}{2*-2}=\frac{-11+\sqrt{281}}{-4}
$