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4x^2-40x+3=0
a = 4; b = -40; c = +3;
Δ = b2-4ac
Δ = -402-4·4·3
Δ = 1552
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{97}}{2*4}=\frac{40-4\sqrt{97}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{97}}{2*4}=\frac{40+4\sqrt{97}}{8} $
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