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4x^2+126x+545=0
a = 4; b = 126; c = +545;
Δ = b2-4ac
Δ = 1262-4·4·545
Δ = 7156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7156}=\sqrt{4*1789}=\sqrt{4}*\sqrt{1789}=2\sqrt{1789}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(126)-2\sqrt{1789}}{2*4}=\frac{-126-2\sqrt{1789}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(126)+2\sqrt{1789}}{2*4}=\frac{-126+2\sqrt{1789}}{8} $
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