45x^2+7x=0

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Solution for 45x^2+7x=0 equation: 



45x^2+7x=0

a = 45; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·45·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*45}=\frac{-14}{90}
=-7/45
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*45}=\frac{0}{90}
=0
$

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