So before we can solve using the quadratic formula, we have to set the equation to zero. We can do that by adding 2 on both sides of the equation [tex] -5y^2 + 2y +2 =0 [/tex]
Now we can use the quadratic formula, which is [tex] y=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a} [/tex] , with a=x^2 coefficient, b=x coefficient, and c=constant. We can form the equation as such: [tex] y=\frac{-2+\sqrt{2^2-4*(-5)*2}}{2*(-5)},\frac{-2-\sqrt{2^2-4*(-5)*2}}{2*(-5)} [/tex]
Firstly, solve the exponents and the multiplications: [tex] y=\frac{-2+\sqrt{4+40}}{-10},\frac{-2-\sqrt{4+40}}{-10} [/tex]
Next, do the addition within the radicals (these will be your exact solutions. but I'll solve for the decimal form as well.): [tex] y=\frac{-2+\sqrt{44}}{-10},\frac{-2-\sqrt{44}}{-10} [/tex]
Next, solve for the rest on your calculator, and your answers will be (rounded to the thousandths) y = -0.463, 0.863
