Question

quadratic equation y^2 -16y-79

Answer 1

see attached pic.
Hope it helps u.

Answer 2

Hey there!

Solving it by applying quadratic formula for equation "y^2 - 16y - 79 = 0"

For a required quadratic equation of $\bf{ay^2 + by + c = 0}$ the solutions for which can be represented by the quadratic formula of :

$\boxed{\bf{y_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = 1,  b = - 16,  c = - 79.

Solving for positive and negative values respectively:

$\bf{y_{1} = \dfrac{- (- 16) - \sqrt{(- 16)^2 - 4(1)(- 79)}}{2(1)}}$

$\bf{y_{1} = \dfrac{16 - \sqrt{256 + 316}}{2 \times 1}}$

$\bf{y_{1} = \dfrac{16 - 2 \sqrt{143}}{2 \times 1}}$

$\bf{y_{1} = \dfrac{2(8 - \sqrt{143})}{2}}$

$\bf{\therefore \quad y_{1} = 8 - \sqrt{143}}$       [Decimal Value = - 3.9583]

For the second solution in Positive form of equation:

$\bf{y_{2} = \dfrac{- (- 16) + \sqrt{(- 16)^2 - 4(1)(- 79)}}{2(1)}}$

$\bf{y_{2} = \dfrac{16 + \sqrt{256 + 316}}{2 \times 1}}$

$\bf{y_{2} = \dfrac{16 + 2 \sqrt{143}}{2 \times 1}}$

$\bf{y_{2} = \dfrac{2(8 + \sqrt{143})}{2}}$

$\bf{\therefore \quad y_{2} = 8 + \sqrt{143}}$      [Decimal Value = 19.9583]

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{y_{1} = 8 - \sqrt{143}}}}$

$\boxed{\underline{\bf{y_{2} = 8 + \sqrt{143}}}}$

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!