(t+1)(t)=5t+7

Solution for (t+1)(t)=5t+7 equation:

(t+1)(t)=5t+7

We move all terms to the left:

(t+1)(t)-(5t+7)=0

We multiply parentheses

t^2+t-(5t+7)=0

We get rid of parentheses

t^2+t-5t-7=0

We add all the numbers together, and all the variables

t^2-4t-7=0

a = 1; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·1·(-7)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{11}}{2*1}=\frac{4-2\sqrt{11}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{11}}{2*1}=\frac{4+2\sqrt{11}}{2}
$