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25x^2+49x-2=0
a = 25; b = 49; c = -2;
Δ = b2-4ac
Δ = 492-4·25·(-2)
Δ = 2601
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2601}=51$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-51}{2*25}=\frac{-100}{50} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+51}{2*25}=\frac{2}{50} =1/25 $
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