What are the solutions to the equation x^2-7x+27=15 ?

The solutions to the equation x^2-7x+27=15 are x=4,x=3

Find the zeros of x^2-7x+27=15

The zeros of x^2-7x+27=15 are x=4,x=3

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x^{2} - 7 x + 27 = 15

x = 4, x = 3

Solution steps

x^{2} - 7 x + 27 = 15

Move

15

to the left side

x^{2} - 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

x_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

x = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

x = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 4, x = 3

- x^{2} - 8 x + 20 = 0

(x - 1)^{2} = \frac{5}{9}

(3 x + 2)^{2} + 4 = (x + 1)^{2}

x, x^{2} - y^{2} + z^{2} = 0

t, f = (t - 5) e t - 5 U (t - 5)

What are the solutions to the equation x^2-7x+27=15 ?
The solutions to the equation x^2-7x+27=15 are x=4,x=3
Find the zeros of x^2-7x+27=15
The zeros of x^2-7x+27=15 are x=4,x=3