Question

X^2-5x-24=0 3x^2+x-4=0 Solve by factoring or using the quadratic formula

Answer 1

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

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

Answer 2

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

$3x^2+x-4=0\\ 3x^2-3x+4x-4=0\\ 3x(x-1)+4(x-1)=0\\ (3x+4)(x-1)=0\\ x=-\frac{4}{3} \vee x=1$