What are the solutions to the equation x^2-17x+52=0 ?

The solutions to the equation x^2-17x+52=0 are x=13,x=4

Find the zeros of x^2-17x+52=0

The zeros of x^2-17x+52=0 are x=13,x=4

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x^{2} - 17 x + 52 = 0

x = 13, x = 4

Solution steps

x^{2} - 17 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

(- 17)^{2} - 4 \cdot 1 \cdot 52 = 9

x_{1, 2} = \frac{- ( - 17 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 17 ) - 9}{2 \cdot 1}

x = \frac{- ( - 17 ) + 9}{2 \cdot 1} : 13

x = \frac{- ( - 17 ) - 9}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 13, x = 4

2 x - 3 (x + 1) \leq 3 x

\frac{8 - x}{9} = \frac{2 x + 1}{3}

75

\frac{( 2 + i )}{( 4 + i )}

\frac{t ^{2}}{20} + 8 = 15

What are the solutions to the equation x^2-17x+52=0 ?
The solutions to the equation x^2-17x+52=0 are x=13,x=4
Find the zeros of x^2-17x+52=0
The zeros of x^2-17x+52=0 are x=13,x=4