5t^2-23+12=0

Solution for 5t^2-23+12=0 equation:

5t^2-23+12=0

We add all the numbers together, and all the variables

5t^2-11=0

a = 5; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·5·(-11)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{55}}{2*5}=\frac{0-2\sqrt{55}}{10}
=-\frac{2\sqrt{55}}{10}
=-\frac{\sqrt{55}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{55}}{2*5}=\frac{0+2\sqrt{55}}{10}
=\frac{2\sqrt{55}}{10}
=\frac{\sqrt{55}}{5}
$