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4x^2+5x-35=0
a = 4; b = 5; c = -35;
Δ = b2-4ac
Δ = 52-4·4·(-35)
Δ = 585
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{65}}{2*4}=\frac{-5-3\sqrt{65}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{65}}{2*4}=\frac{-5+3\sqrt{65}}{8} $
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