What are the solutions to the equation 17x^2-21x=-3x^2-5 ?

The solutions to the equation 17x^2-21x=-3x^2-5 are x=(21+sqrt(41))/(40),x=(21-sqrt(41))/(40)

Find the zeros of 17x^2-21x=-3x^2-5

The zeros of 17x^2-21x=-3x^2-5 are x=(21+sqrt(41))/(40),x=(21-sqrt(41))/(40)

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17 x^{2} - 21 x = - 3 x^{2} - 5

x = \frac{21 + 41}{40}, x = \frac{21 - 41}{40}

Decimal Notation

x = 0.68507 \dots, x = 0.36492 \dots

Solution steps

17 x^{2} - 21 x = - 3 x^{2} - 5

Move

5

to the left side

17 x^{2} - 21 x + 5 = - 3 x^{2}

Move

3 x^{2}

to the left side

20 x^{2} - 21 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 20 \cdot 5}{2 \cdot 20}

(- 21)^{2} - 4 \cdot 20 \cdot 5 = 41

x_{1, 2} = \frac{- ( - 21 ) \pm 41}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 41}{2 \cdot 20}, x_{2} = \frac{- ( - 21 ) - 41}{2 \cdot 20}

x = \frac{- ( - 21 ) + 41}{2 \cdot 20} : \frac{21 + 41}{40}

x = \frac{- ( - 21 ) - 41}{2 \cdot 20} : \frac{21 - 41}{40}

The solutions to the quadratic equation are:

x = \frac{21 + 41}{40}, x = \frac{21 - 41}{40}

(a - x)^{3} + (x - b)^{3} = (a - b)^{3}

x^{2} + 1 = 2 x + 1

3 x^{2} + 6 x + 10 = 0

x^{2} + 15 x + 64 = 20

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

What are the solutions to the equation 17x^2-21x=-3x^2-5 ?
The solutions to the equation 17x^2-21x=-3x^2-5 are x=(21+sqrt(41))/(40),x=(21-sqrt(41))/(40)
Find the zeros of 17x^2-21x=-3x^2-5
The zeros of 17x^2-21x=-3x^2-5 are x=(21+sqrt(41))/(40),x=(21-sqrt(41))/(40)