What are the solutions to the equation 7x^2-18=-39x ?

The solutions to the equation 7x^2-18=-39x are x= 3/7 ,x=-6

Find the zeros of 7x^2-18=-39x

The zeros of 7x^2-18=-39x are x= 3/7 ,x=-6

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7 x^{2} - 18 = - 39 x

x = \frac{3}{7}, x = - 6

Decimal Notation

x = 0.42857 \dots, x = - 6

Solution steps

7 x^{2} - 18 = - 39 x

Move

39 x

to the left side

7 x^{2} - 18 + 39 x = 0

Write in the standard form

a x^{2} + b x + c = 0

7 x^{2} + 39 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 39 \pm 3 9 ^{2} - 4 \cdot 7 ( - 18 )}{2 \cdot 7}

3 9^{2} - 4 \cdot 7 (- 18) = 45

x_{1, 2} = \frac{- 39 \pm 45}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 39 + 45}{2 \cdot 7}, x_{2} = \frac{- 39 - 45}{2 \cdot 7}

x = \frac{- 39 + 45}{2 \cdot 7} : \frac{3}{7}

x = \frac{- 39 - 45}{2 \cdot 7} : - 6

The solutions to the quadratic equation are:

x = \frac{3}{7}, x = - 6

\frac{7 ^{- 2}}{7 ^{- 6}}

(- \frac{1}{5})^{3}

x^{7} + y^{7}

\frac{2 \cdot 0.0 7 ^{2} \cdot 980}{9 \cdot 1 0 ^{- 4} \cdot 1.8}

K (r, s) = 4 r - 7 s

What are the solutions to the equation 7x^2-18=-39x ?
The solutions to the equation 7x^2-18=-39x are x= 3/7 ,x=-6
Find the zeros of 7x^2-18=-39x
The zeros of 7x^2-18=-39x are x= 3/7 ,x=-6