# Solve quadratic equation. 9x^{2}+6x+1=0

Jason Watson 2021-11-21 Answered
$9{x}^{2}+6x+1=0$
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Mespirst
Step 1
Consider the given equation.
$9{x}^{2}+6x+1=0$
Now, find the factor of the equation.
$9{x}^{2}+6x+1=0$
$9{x}^{2}+3x+3x+1=0$
3x(3x+1)+1(3x+1)=0
${\left(3x+1\right)}^{2}=0$
Step 2
Further simplify.
(3x+1)=0
3x=-1
$x=-\frac{1}{3}$
Therefore, the solution of the equation is $x=-\frac{1}{3}$.
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Mespirst
Step 1: Use the formula for the roots of the quadratic equation
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
In the standard form, determine a, b and c from the original equation and insert them into the formula for the roots of the quadratic equation.
$9{x}^{2}+6x+1=0$
a=9
b=6
c=1
$x=\frac{-6±\sqrt{{6}^{2}-4\cdot 9\cdot 1}}{2\cdot 9}$
Step 2: Raise to the degree
$x=\frac{-6±\sqrt{36-4\cdot 9\cdot 1}}{2\cdot 9}$
$x=\frac{-6±\sqrt{36-4\cdot 9\cdot 1}}{2\cdot 9}$
Step 3: Multiply the numbers
$x=\frac{-6±\sqrt{36-4\cdot 9\cdot 1}}{2\cdot 9}$
$x=\frac{-6±\sqrt{36-36}}{2\cdot 9}$
Step 4: Subtract the numbers
$x=\frac{-6±\sqrt{36-36}}{2\cdot 9}$
$x=\frac{-6±\sqrt{0}}{2\cdot 9}$
Step 5: Calculate the square root
$x=\frac{-6±\sqrt{0}}{2\cdot 9}$
$x=\frac{-6±0}{2\cdot 9}$
Step 6: Add zero
$x=\frac{-6±0}{2\cdot 9}$
$x=\frac{-6}{2\cdot 9}$
Step 7: Multiply the numbers
$x=\frac{-6}{2\cdot 9}$
$x=\frac{-6}{18}$
Step 8: Reduce the same terms in the numerator and denominator
$x=\frac{-6}{18}$
$x=-\frac{1}{3}$
The solution
$x=-\frac{1}{3}$