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3.5x^2+125=42x
We move all terms to the left:
3.5x^2+125-(42x)=0
a = 3.5; b = -42; c = +125;
Δ = b2-4ac
Δ = -422-4·3.5·125
Δ = 14
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{-(-42)-\sqrt{14}}{2*3.5}=\frac{42-\sqrt{14}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+\sqrt{14}}{2*3.5}=\frac{42+\sqrt{14}}{7} $
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