(-3t^2+7t-5)+(6t^2+4t+1)=0

Solution for (-3t^2+7t-5)+(6t^2+4t+1)=0 equation:

(-3t^2+7t-5)+(6t^2+4t+1)=0

We get rid of parentheses

-3t^2+6t^2+7t+4t-5+1=0

We add all the numbers together, and all the variables

3t^2+11t-4=0

a = 3; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·3·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*3}=\frac{-24}{6}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*3}=\frac{2}{6}
=1/3
$