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49x^2+14x-18=0
a = 49; b = 14; c = -18;
Δ = b2-4ac
Δ = 142-4·49·(-18)
Δ = 3724
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3724}=\sqrt{196*19}=\sqrt{196}*\sqrt{19}=14\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{19}}{2*49}=\frac{-14-14\sqrt{19}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{19}}{2*49}=\frac{-14+14\sqrt{19}}{98} $
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