4t^2-28t=6t^2-12t+26

Solution for 4t^2-28t=6t^2-12t+26 equation:

4t^2-28t=6t^2-12t+26

We move all terms to the left:

4t^2-28t-(6t^2-12t+26)=0

We get rid of parentheses

4t^2-6t^2-28t+12t-26=0

We add all the numbers together, and all the variables

-2t^2-16t-26=0

a = -2; b = -16; c = -26;
Δ = b2-4ac
Δ = -162-4·(-2)·(-26)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{3}}{2*-2}=\frac{16-4\sqrt{3}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{3}}{2*-2}=\frac{16+4\sqrt{3}}{-4}
$