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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+t^2+6t-4t-24)-5=0

We get rid of parentheses

t^2+6t-4t-24-5=0

We add all the numbers together, and all the variables

t^2+2t-29=0

a = 1; b = 2; c = -29;
Δ = b2-4ac
Δ = 22-4·1·(-29)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{30}}{2*1}=\frac{-2-2\sqrt{30}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{30}}{2*1}=\frac{-2+2\sqrt{30}}{2}
$