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4x^2+41x+98=2-3x
We move all terms to the left:
4x^2+41x+98-(2-3x)=0
We add all the numbers together, and all the variables
4x^2+41x-(-3x+2)+98=0
We get rid of parentheses
4x^2+41x+3x-2+98=0
We add all the numbers together, and all the variables
4x^2+44x+96=0
a = 4; b = 44; c = +96;
Δ = b2-4ac
Δ = 442-4·4·96
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-20}{2*4}=\frac{-64}{8} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+20}{2*4}=\frac{-24}{8} =-3 $
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