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

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

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

We move all terms to the left:

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

We multiply parentheses

-12t^2+20t-(2t-6)+7=0

We get rid of parentheses

-12t^2+20t-2t+6+7=0

We add all the numbers together, and all the variables

-12t^2+18t+13=0

a = -12; b = 18; c = +13;
Δ = b2-4ac
Δ = 182-4·(-12)·13
Δ = 948
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:

$\sqrt{\Delta}=\sqrt{948}=\sqrt{4*237}=\sqrt{4}*\sqrt{237}=2\sqrt{237}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{237}}{2*-12}=\frac{-18-2\sqrt{237}}{-24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{237}}{2*-12}=\frac{-18+2\sqrt{237}}{-24}
$