quadratic equaiton is
x=\(\frac{-b- \sqrt{b^2-4ac} }{2a}\) or \(\frac{-b+ \sqrt{b^2-4ac} }{2a}\)
just subsitute and simplify
I will only do one in detail since it takes some time
so first one2b^2+10b+12
a=2
b=10
c=12
subsitute
x=\(\frac{-10- \sqrt{10^2-4(2)(12)} }{2(2)}\) or
\(\frac{-10+ \sqrt{10^2-4(2)(12)} }{2(2)}\)
pemdas
\(\frac{-10- \sqrt{10^2-4(2)(12)} }{2(2)}\)
do parenthasees or inside exponents
(10^2-4(2)(12) =(100-96)=4
square root of 4=2
(-10-2)/(2 times 2)=-12/4=-3
if -10+square root then
(-10+2)/(2 times 2)=-8/4=-2
answer is
b=-3 or -2
second one
answer is g=-2 or -2/3
3rd
x=-3/2 or 1/2
4th
b=\(\frac{5- \sqrt{345} }{16}\) or
\(\frac{5+ \sqrt{345} }{16}\)
5th
m=-3/2 or 1/3
6th
d=-5/2 or 4/5
