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12x^2+6x+5(2x+1)=0
We multiply parentheses
12x^2+6x+10x+5=0
We add all the numbers together, and all the variables
12x^2+16x+5=0
a = 12; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·12·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*12}=\frac{-20}{24} =-5/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*12}=\frac{-12}{24} =-1/2 $
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