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x^2+22x+48=0
a = 1; b = 22; c = +48;
Δ = b2-4ac
Δ = 222-4·1·48
Δ = 292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{73}}{2*1}=\frac{-22-2\sqrt{73}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{73}}{2*1}=\frac{-22+2\sqrt{73}}{2} $
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