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5x^2+36x+1=0
a = 5; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·5·1
Δ = 1276
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1276}=\sqrt{4*319}=\sqrt{4}*\sqrt{319}=2\sqrt{319}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{319}}{2*5}=\frac{-36-2\sqrt{319}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{319}}{2*5}=\frac{-36+2\sqrt{319}}{10} $
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