5(t+1)(-t+3)=3

Solution for 5(t+1)(-t+3)=3 equation:

5(t+1)(-t+3)=3

We move all terms to the left:

5(t+1)(-t+3)-(3)=0

We add all the numbers together, and all the variables

5(t+1)(-1t+3)-3=0

We multiply parentheses ..

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

We multiply parentheses

-5t^2+15t-5t+15-3=0

We add all the numbers together, and all the variables

-5t^2+10t+12=0

a = -5; b = 10; c = +12;
Δ = b2-4ac
Δ = 102-4·(-5)·12
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{85}}{2*-5}=\frac{-10-2\sqrt{85}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{85}}{2*-5}=\frac{-10+2\sqrt{85}}{-10}
$