What are the solutions to the equation x^2-15x+26=-3x-6 ?

The solutions to the equation x^2-15x+26=-3x-6 are x=8,x=4

Find the zeros of x^2-15x+26=-3x-6

The zeros of x^2-15x+26=-3x-6 are x=8,x=4

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x^{2} - 15 x + 26 = - 3 x - 6

x = 8, x = 4

Solution steps

x^{2} - 15 x + 26 = - 3 x - 6

Move

6

to the left side

x^{2} - 15 x + 32 = - 3 x

Move

3 x

to the left side

x^{2} - 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

x = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

x = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 8, x = 4

8 x^{2} + 24 x - 20 = 0

2 x^{2} - 5 x - 7 = 0

(- x - 1)^{2} = 0

x, f (x yz) = x^{2} + y^{2} + z^{2}

\frac{1}{2} x^{2} + 5 x = - 17

What are the solutions to the equation x^2-15x+26=-3x-6 ?
The solutions to the equation x^2-15x+26=-3x-6 are x=8,x=4
Find the zeros of x^2-15x+26=-3x-6
The zeros of x^2-15x+26=-3x-6 are x=8,x=4