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0.2x^2+14=480+10x
We move all terms to the left:
0.2x^2+14-(480+10x)=0
We add all the numbers together, and all the variables
0.2x^2-(10x+480)+14=0
We get rid of parentheses
0.2x^2-10x-480+14=0
We add all the numbers together, and all the variables
0.2x^2-10x-466=0
a = 0.2; b = -10; c = -466;
Δ = b2-4ac
Δ = -102-4·0.2·(-466)
Δ = 472.8
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{472.8}}{2*0.2}=\frac{10-\sqrt{472.8}}{0.4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{472.8}}{2*0.2}=\frac{10+\sqrt{472.8}}{0.4} $
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