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5y^2-42y-49=0
a = 5; b = -42; c = -49;
Δ = b2-4ac
Δ = -422-4·5·(-49)
Δ = 2744
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2744}=\sqrt{196*14}=\sqrt{196}*\sqrt{14}=14\sqrt{14}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{14}}{2*5}=\frac{42-14\sqrt{14}}{10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{14}}{2*5}=\frac{42+14\sqrt{14}}{10} $
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