2t^2+7=4t^2-10

Solution for 2t^2+7=4t^2-10 equation:

2t^2+7=4t^2-10

We move all terms to the left:

2t^2+7-(4t^2-10)=0

We get rid of parentheses

2t^2-4t^2+10+7=0

We add all the numbers together, and all the variables

-2t^2+17=0

a = -2; b = 0; c = +17;
Δ = b2-4ac
Δ = 02-4·(-2)·17
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{34}}{2*-2}=\frac{0-2\sqrt{34}}{-4}
=-\frac{2\sqrt{34}}{-4}
=-\frac{\sqrt{34}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{34}}{2*-2}=\frac{0+2\sqrt{34}}{-4}
=\frac{2\sqrt{34}}{-4}
=\frac{\sqrt{34}}{-2}
$