(-6t^2+7t-1)+(2t^2+4t+5)=0

Solution for (-6t^2+7t-1)+(2t^2+4t+5)=0 equation:

(-6t^2+7t-1)+(2t^2+4t+5)=0

We get rid of parentheses

-6t^2+2t^2+7t+4t-1+5=0

We add all the numbers together, and all the variables

-4t^2+11t+4=0

a = -4; b = 11; c = +4;
Δ = b2-4ac
Δ = 112-4·(-4)·4
Δ = 185
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{185}}{2*-4}=\frac{-11-\sqrt{185}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{185}}{2*-4}=\frac{-11+\sqrt{185}}{-8}
$