25+2t=5(t2)

Solution for 25+2t=5(t2) equation:

25+2t=5(t2)

We move all terms to the left:

25+2t-(5(t2))=0

determiningTheFunctionDomain
2t-5t2+25=0

We add all the numbers together, and all the variables

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

a = -5; b = 2; c = +25;
Δ = b2-4ac
Δ = 22-4·(-5)·25
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{14}}{2*-5}=\frac{-2-6\sqrt{14}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{14}}{2*-5}=\frac{-2+6\sqrt{14}}{-10}
$