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7x^2+49x=0
a = 7; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·7·0
Δ = 2401
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{2401}=49$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*7}=\frac{-98}{14} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*7}=\frac{0}{14} =0 $
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