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(10x-5)(4x+3)=180
We move all terms to the left:
(10x-5)(4x+3)-(180)=0
We multiply parentheses ..
(+40x^2+30x-20x-15)-180=0
We get rid of parentheses
40x^2+30x-20x-15-180=0
We add all the numbers together, and all the variables
40x^2+10x-195=0
a = 40; b = 10; c = -195;
Δ = b2-4ac
Δ = 102-4·40·(-195)
Δ = 31300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{31300}=\sqrt{100*313}=\sqrt{100}*\sqrt{313}=10\sqrt{313}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{313}}{2*40}=\frac{-10-10\sqrt{313}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{313}}{2*40}=\frac{-10+10\sqrt{313}}{80} $
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