What are the solutions to the equation 2x^2-21x+40=0 ?

The solutions to the equation 2x^2-21x+40=0 are x=8,x= 5/2

Find the zeros of 2x^2-21x+40=0

The zeros of 2x^2-21x+40=0 are x=8,x= 5/2

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2 x^{2} - 21 x + 40 = 0

x = 8, x = \frac{5}{2}

Decimal Notation

x = 8, x = 2.5

Solution steps

2 x^{2} - 21 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 2 \cdot 40}{2 \cdot 2}

(- 21)^{2} - 4 \cdot 2 \cdot 40 = 11

x_{1, 2} = \frac{- ( - 21 ) \pm 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 11}{2 \cdot 2}, x_{2} = \frac{- ( - 21 ) - 11}{2 \cdot 2}

x = \frac{- ( - 21 ) + 11}{2 \cdot 2} : 8

x = \frac{- ( - 21 ) - 11}{2 \cdot 2} : \frac{5}{2}

The solutions to the quadratic equation are:

x = 8, x = \frac{5}{2}

x^{2} - 5.1 x + 6.3 = 0

\frac{x ^{2}}{3} = 4 x - x^{2}

(x - 2) (x - 6) = 5

4 h^{2} - 9 = 0

x, y x^{2} - 5 x^{2} - 3 y + 4 = 0

What are the solutions to the equation 2x^2-21x+40=0 ?
The solutions to the equation 2x^2-21x+40=0 are x=8,x= 5/2
Find the zeros of 2x^2-21x+40=0
The zeros of 2x^2-21x+40=0 are x=8,x= 5/2