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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2t^2+8t-6=0

a = 2; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·2·(-6)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{7}}{2*2}=\frac{-8-4\sqrt{7}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{7}}{2*2}=\frac{-8+4\sqrt{7}}{4}
$