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4r^2+18r+7=0
a = 4; b = 18; c = +7;
Δ = b2-4ac
Δ = 182-4·4·7
Δ = 212
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{53}}{2*4}=\frac{-18-2\sqrt{53}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{53}}{2*4}=\frac{-18+2\sqrt{53}}{8} $
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