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10x^2-49=0
a = 10; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·10·(-49)
Δ = 1960
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1960}=\sqrt{196*10}=\sqrt{196}*\sqrt{10}=14\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{10}}{2*10}=\frac{0-14\sqrt{10}}{20} =-\frac{14\sqrt{10}}{20} =-\frac{7\sqrt{10}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{10}}{2*10}=\frac{0+14\sqrt{10}}{20} =\frac{14\sqrt{10}}{20} =\frac{7\sqrt{10}}{10} $
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