What are the solutions to the equation x^2+3x=24 ?

The solutions to the equation x^2+3x=24 are x=(-3+sqrt(105))/2 ,x=(-3-sqrt(105))/2

Find the zeros of x^2+3x=24

The zeros of x^2+3x=24 are x=(-3+sqrt(105))/2 ,x=(-3-sqrt(105))/2

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x^{2} + 3 x = 24

x = \frac{- 3 + 105}{2}, x = \frac{- 3 - 105}{2}

Decimal Notation

x = 3.62347 \dots, x = - 6.62347 \dots

Solution steps

x^{2} + 3 x = 24

Move

24

to the left side

x^{2} + 3 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 24) = 105

x_{1, 2} = \frac{- 3 \pm 105}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 105}{2 \cdot 1}, x_{2} = \frac{- 3 - 105}{2 \cdot 1}

x = \frac{- 3 + 105}{2 \cdot 1} : \frac{- 3 + 105}{2}

x = \frac{- 3 - 105}{2 \cdot 1} : \frac{- 3 - 105}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 105}{2}, x = \frac{- 3 - 105}{2}

\frac{15 d ^{- 9} t ^{4}}{( 5 d ^{- 3} t ) ^{- 2}}

7 - \frac{2}{3} x < x - 8

\frac{( 4 ) ( 59.58 \cdot 1 0 ^{- 6} )}{π}

\frac{202}{210} + \frac{11}{3}

4^{- 3}

What are the solutions to the equation x^2+3x=24 ?
The solutions to the equation x^2+3x=24 are x=(-3+sqrt(105))/2 ,x=(-3-sqrt(105))/2
Find the zeros of x^2+3x=24
The zeros of x^2+3x=24 are x=(-3+sqrt(105))/2 ,x=(-3-sqrt(105))/2