14t^2+8-18=0

Solution for 14t^2+8-18=0 equation:

14t^2+8-18=0

We add all the numbers together, and all the variables

14t^2-10=0

a = 14; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·14·(-10)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*14}=\frac{0-4\sqrt{35}}{28}
=-\frac{4\sqrt{35}}{28}
=-\frac{\sqrt{35}}{7}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*14}=\frac{0+4\sqrt{35}}{28}
=\frac{4\sqrt{35}}{28}
=\frac{\sqrt{35}}{7}
$