What are the solutions to the equation x^2-14x+48=0 ?

The solutions to the equation x^2-14x+48=0 are x=8,x=6

Find the zeros of x^2-14x+48=0

The zeros of x^2-14x+48=0 are x=8,x=6

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x^{2} - 14 x + 48 = 0

x = 8, x = 6

Solution steps

x^{2} - 14 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 48 = 2

x_{1, 2} = \frac{- ( - 14 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 2}{2 \cdot 1}

x = \frac{- ( - 14 ) + 2}{2 \cdot 1} : 8

x = \frac{- ( - 14 ) - 2}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 8, x = 6

2 x + 10 \leq 26

a^{8} a^{- 4}

8 x + 11 = 8 x + 8

∣ 6 x + 2 ∣ - 15 < - 1

2 t^{2} - 5 t = 3

What are the solutions to the equation x^2-14x+48=0 ?
The solutions to the equation x^2-14x+48=0 are x=8,x=6
Find the zeros of x^2-14x+48=0
The zeros of x^2-14x+48=0 are x=8,x=6