g^2-10g+21=0

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Solution for g^2-10g+21=0 equation: 



g^2-10g+21=0

a = 1; b = -10; c = +21;
Δ = b2-4ac
Δ = -102-4·1·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*1}=\frac{6}{2}
=3
$


$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*1}=\frac{14}{2}
=7
$

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