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q^2+5q-1050=0
a = 1; b = 5; c = -1050;
Δ = b2-4ac
Δ = 52-4·1·(-1050)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4225}=65$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-65}{2*1}=\frac{-70}{2} =-35 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+65}{2*1}=\frac{60}{2} =30 $
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