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16x^2+58x+49=0
a = 16; b = 58; c = +49;
Δ = b2-4ac
Δ = 582-4·16·49
Δ = 228
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-2\sqrt{57}}{2*16}=\frac{-58-2\sqrt{57}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+2\sqrt{57}}{2*16}=\frac{-58+2\sqrt{57}}{32} $
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