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4x^2+22x+1=0
a = 4; b = 22; c = +1;
Δ = b2-4ac
Δ = 222-4·4·1
Δ = 468
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-6\sqrt{13}}{2*4}=\frac{-22-6\sqrt{13}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+6\sqrt{13}}{2*4}=\frac{-22+6\sqrt{13}}{8} $
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