33=8t+(1/2)(2)t^2

Solution for 33=8t+(1/2)(2)t^2 equation:

33=8t+(1/2)(2)t^2

We move all terms to the left:

33-(8t+(1/2)(2)t^2)=0


Domain of the equation: 2)2t^2)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

-(8t+(+1/2)2t^2)+33=0

We multiply all the terms by the denominator


-(8t+(+1+33*2)2t^2)=0


We calculate terms in parentheses: -(8t+(+1+33*2)2t^2), so:

8t+(+1+33*2)2t^2

We add all the numbers together, and all the variables

8t+672t^2

We add all the numbers together, and all the variables

672t^2+8t

Back to the equation:

-(672t^2+8t)


We get rid of parentheses

-672t^2-8t=0

a = -672; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-672)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-672}=\frac{0}{-1344}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-672}=\frac{16}{-1344}
=-1/84
$