What are the solutions to the equation x^2-27x+50=0 ?

The solutions to the equation x^2-27x+50=0 are x=25,x=2

Find the zeros of x^2-27x+50=0

The zeros of x^2-27x+50=0 are x=25,x=2

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x^{2} - 27 x + 50 = 0

x = 25, x = 2

Solution steps

x^{2} - 27 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 1 \cdot 50}{2 \cdot 1}

(- 27)^{2} - 4 \cdot 1 \cdot 50 = 23

x_{1, 2} = \frac{- ( - 27 ) \pm 23}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 27 ) + 23}{2 \cdot 1}, x_{2} = \frac{- ( - 27 ) - 23}{2 \cdot 1}

x = \frac{- ( - 27 ) + 23}{2 \cdot 1} : 25

x = \frac{- ( - 27 ) - 23}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 25, x = 2

2 - 2

(x^{3})^{\frac{2}{3}} y^{3}

625 t^{2} + 100 t^{1} + 4

a^{4} + 4 a^{2} m + 4 m^{2}

\frac{2 ^{3} + 4 ^{2}}{( \frac{2}{3} ) ^{2}}

What are the solutions to the equation x^2-27x+50=0 ?
The solutions to the equation x^2-27x+50=0 are x=25,x=2
Find the zeros of x^2-27x+50=0
The zeros of x^2-27x+50=0 are x=25,x=2