What are the solutions to the equation -3x^2-48x+543=0 ?

The solutions to the equation -3x^2-48x+543=0 are x=-8-7sqrt(5),x=7sqrt(5)-8

Find the zeros of -3x^2-48x+543=0

The zeros of -3x^2-48x+543=0 are x=-8-7sqrt(5),x=7sqrt(5)-8

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- 3 x^{2} - 48 x + 543 = 0

x = - 8 - 75, x = 75 - 8

Decimal Notation

x = - 23.65247 \dots, x = 7.65247 \dots

Solution steps

- 3 x^{2} - 48 x + 543 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 ( - 3 ) \cdot 543}{2 ( - 3 )}

(- 48)^{2} - 4 (- 3) \cdot 543 = 425

x_{1, 2} = \frac{- ( - 48 ) \pm 42 5}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 42 5}{2 ( - 3 )}, x_{2} = \frac{- ( - 48 ) - 42 5}{2 ( - 3 )}

x = \frac{- ( - 48 ) + 42 5}{2 ( - 3 )} : - 8 - 75

x = \frac{- ( - 48 ) - 42 5}{2 ( - 3 )} : 75 - 8

The solutions to the quadratic equation are:

x = - 8 - 75, x = 75 - 8

0.3 \cdot 0.3

64 - x^{2} y^{2}

x - 4 (0) = 8

n^{2} + 10 n + 9

\frac{5}{2} \cdot \frac{3}{2}

What are the solutions to the equation -3x^2-48x+543=0 ?
The solutions to the equation -3x^2-48x+543=0 are x=-8-7sqrt(5),x=7sqrt(5)-8
Find the zeros of -3x^2-48x+543=0
The zeros of -3x^2-48x+543=0 are x=-8-7sqrt(5),x=7sqrt(5)-8