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3x^2+4x+21x+28=0
We add all the numbers together, and all the variables
3x^2+25x+28=0
a = 3; b = 25; c = +28;
Δ = b2-4ac
Δ = 252-4·3·28
Δ = 289
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*3}=\frac{-42}{6} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*3}=\frac{-8}{6} =-1+1/3 $
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