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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+10t^2-30t+8t-24)-4=0

We get rid of parentheses

10t^2-30t+8t-24-4=0

We add all the numbers together, and all the variables

10t^2-22t-28=0

a = 10; b = -22; c = -28;
Δ = b2-4ac
Δ = -222-4·10·(-28)
Δ = 1604
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:

$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{401}}{2*10}=\frac{22-2\sqrt{401}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{401}}{2*10}=\frac{22+2\sqrt{401}}{20}
$