Question

How do you solve these algebra questains?

2m^2-7m-3=0
4b^2+8b+7=4
5√80a^2
-6√150r
2/√3
√6/5√3

Answer 1

`2m^2-7m-3=0\\a=2;\ b=-7;\ c=-3\\\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot2\cdot(-3)=49+24=73\\\\x=(-b\pm\sqrt\Delta)/(2a)\to \boxed{x=(7\pm√(73))/(2\cdot2)=(7\pm√(73))/(4)}`



`4b^2+8b+7=4\\4b^2+8b+7-4=0\\4b^2+8b+3=0\\4b^2+2b+6b+3=0\\2b(2b+1)+3(2b+1)=0\\(2b+1)(2b+3)=0\iff2b+1=0\ or\ 2b+3=0\\2b=-1\ or\ 2b=-3\\\boxed{b=-(1)/(2)\ or\ b=-(3)/(2)}`



`5√(80a^2)=5√(80)\cdot√(a^2)=5√(16\cdot5)\cdot|a|=5\cdot√(16)\cdot\sqrt5\cdot|a|\\\\=5\cdot4\cdot\sqrt5\cdot|a|=\boxed20`



`-6√(150r)=-6√(25\cdot6r)=-6√(25)\cdot√(6r)=-6\cdot5√(6r)=\boxed{-30√(6r)}`



`(2)/(\sqrt3)=(2)/(\sqrt3)\cdot(\sqrt3)/(\sqrt3)=\boxed{(2\sqrt3)/(3)}`



`(\sqrt6)/(5\sqrt3)=(1)/(5)\cdot(\sqrt6)/(\sqrt3)=(1)/(5)\sqrt(6)/(3)=(1)/(5)\sqrt2=\boxed{(\sqrt2)/(5)}`