What are the solutions to the equation x^2-28x+132=0 ?

The solutions to the equation x^2-28x+132=0 are x=22,x=6

Find the zeros of x^2-28x+132=0

The zeros of x^2-28x+132=0 are x=22,x=6

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x^{2} - 28 x + 132 = 0

x = 22, x = 6

Solution steps

x^{2} - 28 x + 132 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 1 \cdot 132}{2 \cdot 1}

(- 28)^{2} - 4 \cdot 1 \cdot 132 = 16

x_{1, 2} = \frac{- ( - 28 ) \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 16}{2 \cdot 1}, x_{2} = \frac{- ( - 28 ) - 16}{2 \cdot 1}

x = \frac{- ( - 28 ) + 16}{2 \cdot 1} : 22

x = \frac{- ( - 28 ) - 16}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 22, x = 6

900 x^{2} + 9998 x + 24330 = 4925.67 (3)

a^{2} - 28 a - 480 = 0

\frac{2}{3} x^{2} - 16 = 32

x, x^{2} + y^{2} - 6 x + 10 y + 34 = 0

3 x^{2} + 4 x = 4

What are the solutions to the equation x^2-28x+132=0 ?
The solutions to the equation x^2-28x+132=0 are x=22,x=6
Find the zeros of x^2-28x+132=0
The zeros of x^2-28x+132=0 are x=22,x=6