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(4.9x^2)+10x-3.5=0
We add all the numbers together, and all the variables
10x+(4.9x^2)-3.5=0
We get rid of parentheses
4.9x^2+10x-3.5=0
a = 4.9; b = 10; c = -3.5;
Δ = b2-4ac
Δ = 102-4·4.9·(-3.5)
Δ = 168.6
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{-(10)-\sqrt{168.6}}{2*4.9}=\frac{-10-\sqrt{168.6}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{168.6}}{2*4.9}=\frac{-10+\sqrt{168.6}}{9.8} $
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