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1x^2+7x-3=0
We add all the numbers together, and all the variables
x^2+7x-3=0
a = 1; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·1·(-3)
Δ = 61
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{61}}{2*1}=\frac{-7-\sqrt{61}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{61}}{2*1}=\frac{-7+\sqrt{61}}{2} $
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