Given quadratic 8x^2+ 2x – 3 ≥ 0.
We know, quadratic formula.
[tex]\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
For given quadratic values of a, b and c are
[tex]a=8,\:b=2,\:c=-3[/tex]
Plugging values in quadratic formula, we get
[tex]x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:8\left(-3\right)}}{2\cdot \:8}[/tex]
[tex]x=\frac{-2+\sqrt{2^2-4\cdot \:8\left(-3\right)}}{2\cdot \:8}[/tex]
[tex]=\frac{-2+10}{16}[/tex]
[tex]=\frac{1}{2}[/tex]
[tex]x=\frac{-2-\sqrt{2^2-4\cdot \:8\left(-3\right)}}{2\cdot \:8}[/tex]
[tex]=\frac{-2-\sqrt{100}}{16}[/tex]
[tex]=\frac{-12}{16}[/tex]
[tex]=-\frac{3}{4}[/tex]
[tex]x=\frac{1}{2},\:x=-\frac{3}{4}[/tex]
Multiplying both sides by in [tex]x=\frac{1}{2}[/tex], we get
[tex]2x=1[/tex]
Subtacting 1 from both sides, we get
2x-1=0
Multiplying by 4 on both sides in [tex]x=-\frac{3}{4}[/tex], we get
[tex]4x=-3[/tex]
Adding 3 on both sides , we get
[tex]4x+3=0[/tex]