0=21x-4.9x^2-18

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Solution for 0=21x-4.9x^2-18 equation:



0=21x-4.9x^2-18
We move all terms to the left:
0-(21x-4.9x^2-18)=0
We add all the numbers together, and all the variables
-(21x-4.9x^2-18)=0
We get rid of parentheses
4.9x^2-21x+18=0
a = 4.9; b = -21; c = +18;
Δ = b2-4ac
Δ = -212-4·4.9·18
Δ = 88.2
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{-(-21)-\sqrt{88.2}}{2*4.9}=\frac{21-\sqrt{88.2}}{9.8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{88.2}}{2*4.9}=\frac{21+\sqrt{88.2}}{9.8} $

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