What are the solutions to the equation q^2-140q+4800=0 ?

The solutions to the equation q^2-140q+4800=0 are q=80,q=60

Find the zeros of q^2-140q+4800=0

The zeros of q^2-140q+4800=0 are q=80,q=60

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q^{2} - 140 q + 4800 = 0

q = 80, q = 60

Solution steps

q^{2} - 140 q + 4800 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 1 \cdot 4800}{2 \cdot 1}

(- 140)^{2} - 4 \cdot 1 \cdot 4800 = 20

q_{1, 2} = \frac{- ( - 140 ) \pm 20}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- ( - 140 ) + 20}{2 \cdot 1}, q_{2} = \frac{- ( - 140 ) - 20}{2 \cdot 1}

q = \frac{- ( - 140 ) + 20}{2 \cdot 1} : 80

q = \frac{- ( - 140 ) - 20}{2 \cdot 1} : 60

The solutions to the quadratic equation are:

q = 80, q = 60

x^{2} + x = 210

x^{2} + 3 x = 2 (x - 1)

0 = y (4 - y)

x, f (h (x)) = x^{2} + 7

x^{2} = 1^{2} + 4^{2}

What are the solutions to the equation q^2-140q+4800=0 ?
The solutions to the equation q^2-140q+4800=0 are q=80,q=60
Find the zeros of q^2-140q+4800=0
The zeros of q^2-140q+4800=0 are q=80,q=60