25=(10(t)^2)/2+5t

Solution for 25=(10(t)^2)/2+5t equation:

25=(10(t)^2)/2+5t

We move all terms to the left:

25-((10(t)^2)/2+5t)=0


Domain of the equation: 2+5t)!=0

We move all terms containing t to the left, all other terms to the right

5t)!=-2

t!=-2/1

t!=-2

t∈R


determiningTheFunctionDomain
-(10t^2/2+5t)+25=0

We get rid of parentheses

-10t^2/2-5t+25=0

We multiply all the terms by the denominator


-10t^2-5t*2+25*2=0

We add all the numbers together, and all the variables

-10t^2-5t*2+50=0

Wy multiply elements

-10t^2-10t+50=0

a = -10; b = -10; c = +50;
Δ = b2-4ac
Δ = -102-4·(-10)·50
Δ = 2100
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:

$\sqrt{\Delta}=\sqrt{2100}=\sqrt{100*21}=\sqrt{100}*\sqrt{21}=10\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{21}}{2*-10}=\frac{10-10\sqrt{21}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{21}}{2*-10}=\frac{10+10\sqrt{21}}{-20}
$