What are the solutions to the equation 4x^2-42x+48=0 ?

The solutions to the equation 4x^2-42x+48=0 are x=(21+sqrt(249))/4 ,x=(21-sqrt(249))/4

Find the zeros of 4x^2-42x+48=0

The zeros of 4x^2-42x+48=0 are x=(21+sqrt(249))/4 ,x=(21-sqrt(249))/4

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4 x^{2} - 42 x + 48 = 0

x = \frac{21 + 249}{4}, x = \frac{21 - 249}{4}

Decimal Notation

x = 9.19493 \dots, x = 1.30506 \dots

Solution steps

4 x^{2} - 42 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 4 \cdot 48}{2 \cdot 4}

(- 42)^{2} - 4 \cdot 4 \cdot 48 = 2249

x_{1, 2} = \frac{- ( - 42 ) \pm 2 249}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 42 ) + 2 249}{2 \cdot 4}, x_{2} = \frac{- ( - 42 ) - 2 249}{2 \cdot 4}

x = \frac{- ( - 42 ) + 2 249}{2 \cdot 4} : \frac{21 + 249}{4}

x = \frac{- ( - 42 ) - 2 249}{2 \cdot 4} : \frac{21 - 249}{4}

The solutions to the quadratic equation are:

x = \frac{21 + 249}{4}, x = \frac{21 - 249}{4}

x^{2} - 17 x + 18 = 0

y^{2} = - 2, 7 + 36 + 8 = 17

139.39 = - 4.9 t^{2} + 73.92 t

2^{3} b^{2} = 2^{7}

4 x^{2} - 16 x + 16 = 4 x

What are the solutions to the equation 4x^2-42x+48=0 ?
The solutions to the equation 4x^2-42x+48=0 are x=(21+sqrt(249))/4 ,x=(21-sqrt(249))/4
Find the zeros of 4x^2-42x+48=0
The zeros of 4x^2-42x+48=0 are x=(21+sqrt(249))/4 ,x=(21-sqrt(249))/4