12t(t+2)=t+10

Solution for 12t(t+2)=t+10 equation:

12t(t+2)=t+10

We move all terms to the left:

12t(t+2)-(t+10)=0

We multiply parentheses

12t^2+24t-(t+10)=0

We get rid of parentheses

12t^2+24t-t-10=0

We add all the numbers together, and all the variables

12t^2+23t-10=0

a = 12; b = 23; c = -10;
Δ = b2-4ac
Δ = 232-4·12·(-10)
Δ = 1009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{1009}}{2*12}=\frac{-23-\sqrt{1009}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{1009}}{2*12}=\frac{-23+\sqrt{1009}}{24}
$