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(2x^2)-(3x)+7=17
We move all terms to the left:
(2x^2)-(3x)+7-(17)=0
We add all the numbers together, and all the variables
2x^2-3x-10=0
a = 2; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·2·(-10)
Δ = 89
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{89}}{2*2}=\frac{3-\sqrt{89}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*2}=\frac{3+\sqrt{89}}{4} $
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