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4s^2+42s+28=0
a = 4; b = 42; c = +28;
Δ = b2-4ac
Δ = 422-4·4·28
Δ = 1316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{329}}{2*4}=\frac{-42-2\sqrt{329}}{8} $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{329}}{2*4}=\frac{-42+2\sqrt{329}}{8} $
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