Question

X^2-5x-24=0

3x^2+x-4=0

Solve by factoring or using the quadratic formula

Answer 1

`x^2-5x-24=0\\ x^2+3x-8x-24=0\\ x(x+3)-8(x+3)=0\\ (x-8)(x+3)=0\\ x=8 \vee x=-3`


`3x^2+x-4=0\\ 3x^2-3x+4x-4=0\\ 3x(x-1)+4(x-1)=0\\ (3x+4)(x-1)=0\\ x=-(4)/(3) \vee x=1`

Answer 2

`x^2-5x-24=0\\\\-5x=3x-8x\\-24=3\cdot(-8)\\\\x^2-5x-24=0\\\\\Downarrow\\\\(x+3)(x-8)=0\iff x+3=0\ \vee\ x-8=0\\\\x=-3\ \vee\ x=8`

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`3x^2+x-4=0\\\\a=3;\ b=1;\ c=-4\\\\\Delta=b^2-4ac\to\Delta=1^2-4\cdot3\cdot(-4)=1+48=49\\\\x_1=(-b-\sqrt\Delta)/(2a)\to x_1=(-1-√(49))/(2\cdot3)=(-1-7)/(6)=(-8)/(6)=-(4)/(3)\\\\x_2=(-b+\sqrt\Delta)/(2a)\to x_2=(-1+√(49))/(2\cdot3)=(-1+7)/(6)=(6)/(6)=1`