What are the solutions to the equation 4x^2-28x+24=0 ?

The solutions to the equation 4x^2-28x+24=0 are x=6,x=1

Find the zeros of 4x^2-28x+24=0

The zeros of 4x^2-28x+24=0 are x=6,x=1

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4 x^{2} - 28 x + 24 = 0

x = 6, x = 1

Solution steps

4 x^{2} - 28 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 24}{2 \cdot 4}

(- 28)^{2} - 4 \cdot 4 \cdot 24 = 20

x_{1, 2} = \frac{- ( - 28 ) \pm 20}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 20}{2 \cdot 4}, x_{2} = \frac{- ( - 28 ) - 20}{2 \cdot 4}

x = \frac{- ( - 28 ) + 20}{2 \cdot 4} : 6

x = \frac{- ( - 28 ) - 20}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 6, x = 1

(x + 4) (x + 9) (5 - x) \leq 0

\frac{1}{3} x^{2} - \frac{1}{9} x - \frac{10}{9} = 0

(- 3 + 6 i)^{2}

9 x^{2} + 12 x - 5

What are the solutions to the equation 4x^2-28x+24=0 ?
The solutions to the equation 4x^2-28x+24=0 are x=6,x=1
Find the zeros of 4x^2-28x+24=0
The zeros of 4x^2-28x+24=0 are x=6,x=1