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x^2+56x+49=0
a = 1; b = 56; c = +49;
Δ = b2-4ac
Δ = 562-4·1·49
Δ = 2940
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2940}=\sqrt{196*15}=\sqrt{196}*\sqrt{15}=14\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-14\sqrt{15}}{2*1}=\frac{-56-14\sqrt{15}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+14\sqrt{15}}{2*1}=\frac{-56+14\sqrt{15}}{2} $
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