Question

Solve the following equation using quadratic formula
${y}^{2} - 14y - 12 = 0$

Answer 1

Hello,
Hello,
y²+14y-12=0
∆=14²-4•(-12)=196+48=244
y₁,₂=14± √∆/2=14±√244/2=14±2√61/2
then
y₁=14-2√61/2=2(7-√61)/2=7-√61
and
y₂=14+2√61/2=2(7+√61)/2=7+√61

therefore
y=7-√61 ,y=7+√61

Bye :-)

Answer 2

Given y^2 - 14y - 12 = 0.

a = 1, b = -14, c = -12.

 For a quadratic equation of the form ax^2 + bx + c = 0 the solutions are 

(1) -b + root b^2 - 4ac/2a.

    = -(-14) + root (-14)^2 - 4 * 1*(-12)/(2 * 1)

    = 14 + root(-14)^2 - 4 * 1 * (-12)/(2)

    = 14 + root 244/(2)

    = 14 + 2 root 61/(2)

    = 2(7 + root 61)(2)

    = 7 + root 61.


(2) b - root b^2 - 4ac/2a

     = -(-14) - root (-14)^2 - 4 * 1 * (-12)/(2)

     = 14 - root (-14)^2- 4 * 1 * (-12)/2

     = 14 - 2 root 61/2

     = 7 - root 61.


The solutions are 7 + root 61 and 7 - root 61.


Hope this helps!