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4y^2-28y+28=0
a = 4; b = -28; c = +28;
Δ = b2-4ac
Δ = -282-4·4·28
Δ = 336
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{21}}{2*4}=\frac{28-4\sqrt{21}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{21}}{2*4}=\frac{28+4\sqrt{21}}{8} $
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