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2x^2+3x+3=7
We move all terms to the left:
2x^2+3x+3-(7)=0
We add all the numbers together, and all the variables
2x^2+3x-4=0
a = 2; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·2·(-4)
Δ = 41
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{41}}{2*2}=\frac{-3-\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*2}=\frac{-3+\sqrt{41}}{4} $
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