What are the solutions to the equation x^2-19x+48=0 ?

The solutions to the equation x^2-19x+48=0 are x=16,x=3

Find the zeros of x^2-19x+48=0

The zeros of x^2-19x+48=0 are x=16,x=3

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x^{2} - 19 x + 48 = 0

x = 16, x = 3

Solution steps

x^{2} - 19 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 19)^{2} - 4 \cdot 1 \cdot 48 = 13

x_{1, 2} = \frac{- ( - 19 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 19 ) - 13}{2 \cdot 1}

x = \frac{- ( - 19 ) + 13}{2 \cdot 1} : 16

x = \frac{- ( - 19 ) - 13}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 16, x = 3

r, A = 2 π r (r + h)

3 w^{2} + 15 w + 12 = w

3 y^{2} + 6 y - 9 = 0

x^{2} + x + 1 = 0

y^{2} - \frac{1}{3} a y = 0

What are the solutions to the equation x^2-19x+48=0 ?
The solutions to the equation x^2-19x+48=0 are x=16,x=3
Find the zeros of x^2-19x+48=0
The zeros of x^2-19x+48=0 are x=16,x=3