1/2t^2-49=0

Solution for 1/2t^2-49=0 equation:

1/2t^2-49=0


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

t^2!=0/2

t^2!=√0

t!=0

t∈R


We multiply all the terms by the denominator


-49*2t^2+1=0

Wy multiply elements

-98t^2+1=0

a = -98; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-98)·1
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{2}}{2*-98}=\frac{0-14\sqrt{2}}{-196}
=-\frac{14\sqrt{2}}{-196}
=-\frac{\sqrt{2}}{-14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{2}}{2*-98}=\frac{0+14\sqrt{2}}{-196}
=\frac{14\sqrt{2}}{-196}
=\frac{\sqrt{2}}{-14}
$