4t-2+5^2=6+t^2

Solution for 4t-2+5^2=6+t^2 equation:

4t-2+5^2=6+t^2

We move all terms to the left:

4t-2+5^2-(6+t^2)=0

We add all the numbers together, and all the variables

-(6+t^2)+4t+23=0

We get rid of parentheses

-t^2+4t-6+23=0

We add all the numbers together, and all the variables

-1t^2+4t+17=0

a = -1; b = 4; c = +17;
Δ = b2-4ac
Δ = 42-4·(-1)·17
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{21}}{2*-1}=\frac{-4-2\sqrt{21}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{21}}{2*-1}=\frac{-4+2\sqrt{21}}{-2}
$