What are the solutions to the equation 4x^2-32x+60=0 ?

The solutions to the equation 4x^2-32x+60=0 are x=5,x=3

Find the zeros of 4x^2-32x+60=0

The zeros of 4x^2-32x+60=0 are x=5,x=3

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4 x^{2} - 32 x + 60 = 0

x = 5, x = 3

Solution steps

4 x^{2} - 32 x + 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 4 \cdot 60}{2 \cdot 4}

(- 32)^{2} - 4 \cdot 4 \cdot 60 = 8

x_{1, 2} = \frac{- ( - 32 ) \pm 8}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 8}{2 \cdot 4}, x_{2} = \frac{- ( - 32 ) - 8}{2 \cdot 4}

x = \frac{- ( - 32 ) + 8}{2 \cdot 4} : 5

x = \frac{- ( - 32 ) - 8}{2 \cdot 4} : 3

The solutions to the quadratic equation are:

x = 5, x = 3

16 x^{2} + 7 = 0

(x - 4)^{2} - 5 (x - 4) = 24

a^{2} - 2 a - 8 = 0

- 20 = 10 t - \frac{9.81 t ^{2}}{2}

30 + 19 x - 4 x^{2} = 0

What are the solutions to the equation 4x^2-32x+60=0 ?
The solutions to the equation 4x^2-32x+60=0 are x=5,x=3
Find the zeros of 4x^2-32x+60=0
The zeros of 4x^2-32x+60=0 are x=5,x=3