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4x^2+62x-270=0
a = 4; b = 62; c = -270;
Δ = b2-4ac
Δ = 622-4·4·(-270)
Δ = 8164
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8164}=\sqrt{4*2041}=\sqrt{4}*\sqrt{2041}=2\sqrt{2041}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-2\sqrt{2041}}{2*4}=\frac{-62-2\sqrt{2041}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+2\sqrt{2041}}{2*4}=\frac{-62+2\sqrt{2041}}{8} $
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