0=-2t2+-9t+56

Solution for 0=-2t2+-9t+56 equation:

0=-2t^2+-9t+56

We move all terms to the left:

0-(-2t^2+-9t+56)=0

We add all the numbers together, and all the variables

-(-2t^2+-9t+56)=0

We use the square of the difference formula

-(-2t^2-9t+56)=0

We get rid of parentheses

2t^2+9t-56=0

a = 2; b = 9; c = -56;
Δ = b2-4ac
Δ = 92-4·2·(-56)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-23}{2*2}=\frac{-32}{4}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+23}{2*2}=\frac{14}{4}
=3+1/2
$