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7x^2+3x+(-5)=0
We add all the numbers together, and all the variables
7x^2+3x-5=0
a = 7; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·7·(-5)
Δ = 149
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{-(3)-\sqrt{149}}{2*7}=\frac{-3-\sqrt{149}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{149}}{2*7}=\frac{-3+\sqrt{149}}{14} $
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