298=2t^2+48

Solution for 298=2t^2+48 equation:

298=2t^2+48

We move all terms to the left:

298-(2t^2+48)=0

We get rid of parentheses

-2t^2-48+298=0

We add all the numbers together, and all the variables

-2t^2+250=0

a = -2; b = 0; c = +250;
Δ = b2-4ac
Δ = 02-4·(-2)·250
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*-2}=\frac{0-20\sqrt{5}}{-4}
=-\frac{20\sqrt{5}}{-4}
=-\frac{5\sqrt{5}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*-2}=\frac{0+20\sqrt{5}}{-4}
=\frac{20\sqrt{5}}{-4}
=\frac{5\sqrt{5}}{-1}
$