7y^2-18y+3=0

Solution for 7y^2-18y+3=0 equation:

7y^2-18y+3=0

a = 7; b = -18; c = +3;
Δ = b2-4ac
Δ = -182-4·7·3
Δ = 240
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{15}}{2*7}=\frac{18-4\sqrt{15}}{14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{15}}{2*7}=\frac{18+4\sqrt{15}}{14}
$