12(t)=-10t^2+22t+8

Solution for 12(t)=-10t^2+22t+8 equation:

12(t)=-10t^2+22t+8

We move all terms to the left:

12(t)-(-10t^2+22t+8)=0

We get rid of parentheses

10t^2-22t+12t-8=0

We add all the numbers together, and all the variables

10t^2-10t-8=0

a = 10; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·10·(-8)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{105}}{2*10}=\frac{10-2\sqrt{105}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{105}}{2*10}=\frac{10+2\sqrt{105}}{20}
$