3/2t2=4

Solution for 3/2t2=4 equation:

3/2t^2=4

We move all terms to the left:

3/2t^2-(4)=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


-4*2t^2+3=0

Wy multiply elements

-8t^2+3=0

a = -8; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-8)·3
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-8}=\frac{0-4\sqrt{6}}{-16}
=-\frac{4\sqrt{6}}{-16}
=-\frac{\sqrt{6}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-8}=\frac{0+4\sqrt{6}}{-16}
=\frac{4\sqrt{6}}{-16}
=\frac{\sqrt{6}}{-4}
$