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2x^2+5x+7=20
We move all terms to the left:
2x^2+5x+7-(20)=0
We add all the numbers together, and all the variables
2x^2+5x-13=0
a = 2; b = 5; c = -13;
Δ = b2-4ac
Δ = 52-4·2·(-13)
Δ = 129
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{-(5)-\sqrt{129}}{2*2}=\frac{-5-\sqrt{129}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{129}}{2*2}=\frac{-5+\sqrt{129}}{4} $
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