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4x^2+16x-46=0
a = 4; b = 16; c = -46;
Δ = b2-4ac
Δ = 162-4·4·(-46)
Δ = 992
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{62}}{2*4}=\frac{-16-4\sqrt{62}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{62}}{2*4}=\frac{-16+4\sqrt{62}}{8} $
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